User:Series and sum

User:Integrals123

$$\sum_{j=0}^{n}\sum_{k=0}^{n}{n-j \choose j}{n-k \choose k}=F_n^2$$

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$$\sum_{j=0}^{n}(-1)^j{n \choose j}\frac{(aj+b)^n}{cj+d}=\frac{(-1)^{n-1}n!\begin{vmatrix} a & b \\c & d \end{vmatrix}^n}{\prod_{k=0}^{n}(kc+d)}$$

$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}x^j{n-j \choose j}\left(\frac{aj+b}{cj+d}\right)^nB_jB_{n-j}=x\frac{(n-1)B_{n-1}}{2}\left(\frac{a+b}{c+d}\right)^n$$

$$\sum_{j=0}^{n}x^j{n-j \choose j}^sB_jB_{n-j}=x\frac{(n-1)^sB_{n-1}}{2}$$

$$2\sum_{j=0}^{n}(-1)^{n-j}{n-1\choose j}(aj+b)^n=-n!a^{n-1}[a(n-1)+2b]^n$$

$$\sum_{j=0}^{n}(-1)^{n-j}{n\choose j}(j+a)^n=n!$$

$$\sum_{j=0}^{n}(-1)^{n-j}{n+1\choose j}(aj+b)^n j$$

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$$\sum_{j=0}^{\lfloor n/2 \rfloor}(-1)^j{n-j \choose j}3^{n-2j}=F_{2n+2}$$

$$\sum_{j=0}^{\lfloor n/2 \rfloor}(-1)^j{n-j \choose j}4^{n-2j}=\frac{Q^{n+1}-Q^{-n-1}}{2\sqrt{3}}$$

$$Q=2+\sqrt{3}$$ -

$$\sum_{i=0}^{n}(-1)^ii{n^{2k}\choose i}=(-1)^n\frac{n}{n^{2k-1}-n^{2k-2}+\cdots-1}{n^{2k} \choose n+1}$$

$$\sum_{i=0}^{n}(-1)^ii{n^{2k-1}\choose i}=(-1)^n\frac{n(n+1)}{n^{2k-2}+n^{2k-3}+\cdots+1}{n^{2k-1} \choose n+1}$$

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$$n=2k-1$$, $$k \ge 1$$

$$\sum_{j=0}^{n}(-1)^j{n \choose j}^2j=n{n-1 \choose \frac{n-1}{2}}$$

$$\lim_{n \to \infty}\sum_{j=0}^{n}\frac{[n(n+1)]^j}=\frac{e+e^{-1}}{2}$$

$$\sum_{j=0}^{n}{n+j \choose 2j}[a(a+1)]^{n-j}=\frac{a^{2n+1}+(a+1)^{2n+1}}{2a+1}$$

$$\sum_{j=0}^{n}{n+j \choose 2j}=F_{2n+1}$$

$$\sum_{j=0}^{n}{n+j \choose 2j}(-4)^j=(-1)^n(2n+1)$$

$$\sum_{j=0}^{n}{n+j \choose 2j}5^j=\frac{\phi^{-4n}}{1+\phi^4}+\frac{\phi^{4n}}{1+\phi^{-4}}$$

$$\sum_{j=0}^{n}{n+j \choose 2j}(-5)^j=(-1)^n(\phi^{2n+1}-\phi^{-2n-1})$$

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$$\sum_{j=1}^{n}{2n+1\choose 2j}{2n+2j-1 \choose 2j-1}B_{2j}=\frac{2n+1}{2}$$

$$\sum_{j=1}^{n}{2n-1\choose 2j}{2n+2j-1 \choose 2j}B_{2j}=(2n+1)(n-1)$$

$$\sum_{j=1}^{n}{n-1\choose j}{n+j-1 \choose j}B_{2j}=-1$$

$$\sum_{j=0}^{n}{6n-9 \choose 6j+6}B_{6j+6}=2(n-2)$$

$$\sum_{j=2}^{n-2}B_jB_{n-j}{n \choose j}=-(n+1)B_n$$

$$\sum_{j=2}^{n-2}jB_jB_{n-j}{n \choose j}=-\frac{n}{2}(n+1)B_n$$

$$\sum_{j=1}^{n-1}j\frac{B_{2j}B_{2n-2j}}{(2j)!(2n-2j)!}=-\frac{n}{2}(2n+1)\frac{B_{2n}}{(2n)!}$$

$$\sum_{j=1}^{n-1}\frac{B_{j}B_{n-j}}{j!(n-j)!}=(n+1)\frac{B_{n}}{n!}$$

$$\sum_{j=1}^{n-1}j\frac{B_{j}B_{n-j}}{j!(n-j)!}=\frac{n}{2}(n+1)\frac{B_{n}}{n!}$$

$$\sum_{j=2}^{n-2}\frac{B_jB_{n-j}}{j!(n-j)!}=\frac{(n+1)B_n}{n!}$$

$$\sum_{j=2}^{n-2}j\frac{B_jB_{n-j}}{j!(n-j)!}=\frac{n(n+1)}{2}\cdot\frac{B_n}{n!}$$

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$$-1\le x \le n$$

$$\sum_{j=1}^{n}2^{2j}B_{2j}{2n-x \choose 2j}=2n-x-1$$

-

$$\sum_{j=1}^{n}\frac{s(n,j)}{2^j}=-\frac{(2n-3)!!}{2^n}$$

$$\sum_{j=1}^{n}s(n,j)B_j=-\frac{n!}{n+1}(-1)^n$$

$$\sum_{j=1}^{n}(-1)^js(n,j)B_j=-\frac{n!}{n(n+1)}(-1)^n$$

$$\sum_{j=1}^{n}s(n-1,j)B_{j-1}=-\frac{n!}{n^2}(-1)^n$$

$$\sum_{j=1}^{n}(-1)^js(n-1,j-1)B_{j-1}=-\frac{n!}{n^2(n-1)}(-1)^n$$

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$$n=2k$$, $$k\ge0$$

$$\left\lfloor\frac{B_{2n}B_{2n+4}}{B_{2n+2}^2}\cdot (n+1)\left(n+\frac{1}{2}\right)\right\rfloor=(n+2)\left(n+\frac{3}{2}\right)$$

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$$\lim_{n \to \infty}\frac{B_{2n+2j-2}}{B_{2n+2j}}\cdot(n+j)\left(n+j-\frac{1}{2}\right)=-\pi^2$$

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$$n=2k$$, $$k\ge0$$

$$\sum_{j=1}^{n}(2^{-j}-2^j){n \choose j-1}\frac{B_j}{j}=\frac{1}{2^n}$$

$$\sum_{j=1}^{n}\left(\frac{2}{3}\right)^j{n \choose j-1}\frac{B_j}{j}=\frac{2}{3^{n+1}}-\frac{1}{n+1}$$

$$\sum_{j=1}^{n}\left(\frac{4}{3}\right)^j{n \choose j-1}\frac{B_j}{j}=\frac{E_n}{3^{n+1}}-\frac{1}{n+1}$$

$$\sum_{j=1}^{n}{n \choose j-1}\frac{B_j}{j\cdot a^j}=\frac{1}{a^{n+1}}\sum_{k=1}^{a-1}k^n-\frac{1}{n+1}$$

$$\sum_{k=0}^{n}{n \choose k}{m \choose k}\frac{k}{(k+1)^2}=\frac{1}{(m+1)(n+1)}+\frac{mn-1}{(m+1)^2(n+1)^2}\cdot\frac{1}{B(m+1,n+1)}$$

$$\sum_{k=0}^{n}{n \choose k}{1/n \choose k}\frac{k}{(k+1)^2}=\frac{n}{(n+1)^2}$$

$$\sum_{k=0}^{n}{n \choose k}^2\frac{k}{(k+1)^2}=\frac{1}{(n+1)^2}+\frac{n-1}{(n+1)^3}\cdot\frac{1}{B(n+1,n+1)}$$

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$$\sum_{k=1}^{n}\frac{2^k k}{2n-k}{2n-k \choose n}={2n \choose n}$$

$$\sum_{k=1}^{n}\frac{2^k}{2n-k}{2n-k \choose n}=\frac{2^{2n}}{2n}$$

$$\sum_{j=0}^{n}{n \choose j}^2{m+n+j\choose 2n}=\frac{1}{B^2(m+1,n+1)}$$

$$\sum_{j=0}^{n}\frac{{n \choose j}^2{m+n+j\choose 2n}}{j+1}=\frac{(n^2+n+m+2mn)(m+n-1)!(m+n)!}{m!^2(n+1)!^2}$$

$$\sum_{j=0}^{n}{n \choose j}^2{m+n+j\choose 2n}\frac{j}{j+1}=\frac{n^2(m+n-1)!(m+n+1)!}{m!^2(n+1)!^2}$$

$$\lim_{n \to \infty}\sum_{k=0}^{n}\frac{n^{2k+1}}\cdot\frac{a^{2k+2}}{2k+2}=\frac{e^a+e^{-a}-2}{2}$$

$$\lim_{n \to \infty}\sum_{k=0}^{n}\frac{n^k}\frac{a^{k+1}}{k+1}=e^a-1$$

$$\lim_{n \to \infty}\sum_{k=0}^{n}\frac{n^k}\frac{a^{k+1}}{k+1}k=e^a(a-1)+1$$

$$\lim_{n \to \infty}\sum_{k=0}^{n}\frac{n^k}\cdot\frac{k^2}{k+1}=e-1$$

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$$\sum_{j=0}^{n}j(2-4^j){2n \choose 2j}B_{2j}=2^{2n}nB_{2n}$$

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$$n=2k$$, $$k\ge0$$

$$\sum_{j=0}^{n}(1-2^j){n \choose j}B_j=(1-2^n)B_n$$

$$\sum_{j=0}^{n}(2-4^j){n \choose j}B_j=2^nB_n$$

$$\sum_{j=0}^{n}(1-2^j+4^j){n \choose j}B_j=B_n$$

$$\sum_{j=0}^{n}(1+2^j-4^j){n \choose j}B_j=B_n$$

$$\sum_{j=0}^{n}4^j{n \choose j}B_j=(2-2^n)B_n$$

$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}{n \choose n-j}\frac{B_jB_{n-j}}{j!(n-j)!}=(-1)^{k+1}n\frac{B_{n-1}}{(n-1)!}$$

general:

$$\sum_{j=0}^{n}{n \choose n-j}^s\frac{B_jB_{n-j}}{j!(n-j)!}=(-1)^{k+1} n^{s}\frac{B_{n-1}}{(n-1)!}$$

$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}\frac{1}{2^j}{n \choose n-j}\frac{B_jB_{n-j}}{j!(n-j)!}=(-1)^{k+1}\frac{[2^{n-2}+1]}{2^n}\cdot \frac{n B_{n-1}}{(n-1)!}$$

$$\sum_{j=0}^{n}\frac{j}{2^j}{n \choose n-j}\frac{B_jB_{n-j}}{j!(n-j)!}=(-1)^{k+1}\frac{[2^{n-2}+n-1]}{2^n}\cdot \frac{n B_{n-1}}{(n-1)!}$$

general:

$$\sum_{j=0}^{n}\frac{j^s}{2^j}{n \choose n-j}\frac{B_jB_{n-j}}{j!(n-j)!}=(-1)^{k+1}\frac{[2^{n-2}+(n-1)^s]}{2^n}\cdot \frac{n B_{n-1}}{(n-1)!}$$

$$\sum_{j=0}^{n}\frac{j^s}{2^j}{n \choose n-j}^t\frac{B_jB_{n-j}}{j!(n-j)!}=(-1)^{k+1}\frac{[2^{n-2}+(n-1)^s]}{2^n}\cdot \frac{n^t B_{n-1}}{(n-1)!}$$

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$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}\frac{1}{2^j}{n \choose n-j}B_jB_{n-j}=(-1)^{k+1}\frac{2^{n-2}+1}{2^n}\cdot n B_{n-1}$$

$$\sum_{j=0}^{n}\frac{j}{2^j}{n \choose n-j}B_jB_{n-j}=(-1)^{k+1}\frac{2^{n-2}+n-1}{2^n}\cdot n B_{n-1}$$

general:

$$\sum_{j=0}^{n}\frac{j^s}{2^j}{n \choose n-j}B_jB_{n-j}=(-1)^{k+1}\frac{2^{n-2}+(n-1)^s}{2^n}\cdot n B_{n-1}$$

$$\sum_{j=0}^{n}\frac{j^s}{2^j}{n \choose n-j}^t B_jB_{n-j}=(-1)^{k+1}\frac{2^{n-2}+(n-1)^s}{2^n}\cdot n^t B_{n-1}$$

$$n=2k$$, $$k\ge0$$

$$\sum_{j=0}^{n}2^j{n \choose n-j}B_{n-j}=n+B_n$$

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$$n\ge0$$

$$\sum_{j=0}^{n}{n \choose n-j}B_{n-j}=0$$

$$\sum_{j=0}^{n}j{n \choose n-j}B_{n-j}=nB_{n-1}$$

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$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}j^2{n \choose n-j}B_{n-j}=nB_{n-1}$$

$$\sum_{j=0}^{n}2^j{n \choose n-j}B_{n-j}=n$$

$$\sum_{j=0}^{n}2^j{n \choose j}j!(n-j)!=n!(2^{n+1}-1)$$

$$\sum_{j=0}^{n}jB_{n-j}{n \choose j}=0$$

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$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}\frac{j}{2^j}{n \choose n-j}B_{n-j}=(-1)^k\cdot\frac{2^{n-2}-1}{2^{n-1}}\cdot nB_{n-1}$$

$$\sum_{j=0}^{n}\frac{j^2}{2^j}{n \choose n-j}B_{n-j}=(-1)^k\cdot\frac{2^{n-2}-1}{2^{n-1}}\cdot nB_{n-1}$$

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$$n=2k$$, $$k\ge0$$

$$\sum_{j=0}^{n}\frac{j}{2^j}{n \choose n-j}B_{n-j}=0$$

$$n=2k$$, $$k\ge2$$

$$\sum_{j=0}^{n}2^jj{n \choose n-j}B_{n-j}=2n(n-1)$$

$$n\ge0$$

$$\sum_{k=0}^{n}B_{n-k}{n \choose k}\frac{(2k)!!}{k!}=n+B_n$$

$$n=2k$$, $$k\ge0$$

$$\sum_{k=0}^{n}B_{n-k}{n \choose k}\frac{(2k)!!}{2^kk!}=B_n$$

$$n=2k$$, $$k\ge0$$

$$\sum_{k=0}^{n}B_{n-k}{n \choose k}\frac{(2k)!!}{k!}\frac{k}{2^k}=0$$

$$n=2k-1$$, $$k\ge1$$

$$\sum_{k=0}^{n}B_{n-k}{n \choose k}\frac{(2k)!!}{k!}\frac{k}{2^k}=nB_{n-1}$$

$$\sum_{k=0}^{n}B_{n-k}{n \choose k}\frac{(2k)!!}{k!}\frac{k^2}{2^k}=nB_{n-1}$$

$$\sum_{k=0}^{n}B_{n-k}{n \choose k}\frac{(2k)!!}{k!}=n$$

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$$(-1)^n\sum_{k=0}^{n}(2)^k{n+k \choose 2k}\frac{(2n-1)!!}{k!}=\sum_{k=0}^{n}(-4)^k{n+k \choose 2k}\frac{(2n-1)!!}{k!}$$

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$$\sum_{k=2}^{2n}(-1)^k{2n+k \choose 2k}\frac{(2n-1)!!}{(k-1)!}=n(2n+1)$$

$$\sum_{k=2}^{2n}(-2)^k{2n+k \choose 2k}\frac{(2n-1)!!}{(k-1)!}=4n(2n+1)$$

$$\sum_{k=0}^{2n}(-1)^k{2n+k \choose 2k}\frac{(2n-1)!!}{k!}=(-1)^n\frac{(2n-1)!!}{(2n)!!}$$

$$\sum_{k=0}^{n}(-2)^k{n+k \choose 2k}\frac{(2n-1)!!}{k!}=(-1)^n$$

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$$n=2k$$, $$k\ge0$$

$$\sum_{j=0}^{n}j{n \choose j}\frac{B_{n-j+1}}{n-j+1}=B_n-1$$

$$\sum_{j=0}^{n}j{n \choose j}\frac{B_{n-j+2}}{n-j+2}=\frac{1}{n+1}-B_n$$

$$\sum_{j=0}^{n}j^2{n \choose j}\frac{B_{n-j+1}}{n-j+1}=B_n-1-n$$

$$\sum_{j=0}^{n}{n \choose j}\frac{B_{n-j+1}}{2^j(n-j+1)}=-\frac{1}{2^{n+1}(n+1)}$$

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$$n=2k-1$$, $$k\ge1$$

$$\sum_{j=0}^{n}j{n \choose j}\frac{B_{n-j+1}}{n-j+1}=-1$$

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$$\sum_{k=0}^{n}{n \choose k}m^{-ak}(m+1)^{b(n-k)}=m^{-an}\left(m+\sum_{j=0}^{b}{b \choose j}m^{b+j}\right)^n$$

$$\sum_{k=0}^{n}k{n \choose k}m^{-ak}(m+1)^{b(n-k)}=nm^{-an}\left(m+\sum_{j=0}^{b}{b \choose j}m^j\right)^{n-1}$$

$$\sum_{k=0}^{n}k^2{n \choose k}m^{-ak}(m+1)^{b(n-k)}=nm^{-an}\left(m+\sum_{j=0}^{b}{b \choose j}m^j\right)^{n-2}\left(n-1+\sum_{j=0}^{b}{b \choose j}m^j\right)$$

$$\sum_{k=0}^{n}{n \choose k}m^k(m+1)^{b(n-k)}=\left(m+\sum_{j=0}^{b}{b \choose j}m^{b+j}\right)^n$$

$$\sum_{k=0}^{n}k{n \choose k}m^k(m+1)^{b(n-k)}=nm\left(m+\sum_{j=0}^{b}{b \choose j}m^j\right)^{n-1}$$

$$\sum_{k=0}^{n}k^2{n \choose k}m^k(m+1)^{b(n-k)}=nm\left(m+\sum_{j=0}^{b}{b \choose j}m^j\right)^{n-2}\left(nm+\sum_{j=0}^{b}{b \choose j}m^j\right)$$

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$$\sum_{k=0}^{n}{n \choose k}m^k(m-1)^{b(n-k)}=\left((-1)^{b+1}m+\sum_{j=0}^{b}(-1)^{j+b}{b \choose j}m^{b-j}\right)^n$$

$$\sum_{k=0}^{n}k{n \choose k}m^k(m-1)^{b(n-k)}=mn\left((-1)^{b+1}m+\sum_{j=0}^{b}(-1)^{j}{b \choose j}m^{b-j}\right)^{n-1}$$

$$\sum_{k=0}^{n}k^2{n \choose k}m^k(m+1)^{b(n-k)}=nm\left((-1)^{b+1}m+\sum_{j=0}^{b}{b \choose j}m^j\right)^{n-2}\left((-1)^{b+1}nm+\sum_{j=0}^{b}{b \choose j}m^j\right)$$

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$$n=2k$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{2^{n-j+2}}{(n-j+1)(n-j+2)}=\frac{1}{n+1}$$

$$\sum_{j=0}^{n}B_j{n \choose j}j\cdot\frac{2^{n-j+2}}{(n-j+1)(n-j+2)}=-\frac{n}{n+1}$$

$$\sum_{j=0}^{n}(-1)^jB_j{n \choose j}\frac{z^{n-j+1}}{n-j+1}=1+2^n+3^n+\cdots +z^n$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{z^{n-j+1}}{n-j+1}=1+2^n+3^n+\cdots +(z-1)^n$$

$$n=2k$$

$$\sum_{j=0}^{n}(-1)^jB_j{n \choose j}\frac{1}{(n-j+1)(n-j+2)}=\frac{1}{n+1}$$

$$\sum_{j=0}^{n}(-1)^jB_j{n \choose j}\frac{j}{(n-j+1)(n-j+2)}=\frac{1}{n+1}$$

$$n=2k$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{j}{n-j+1}=-B_n$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{j}{n-j+2}=-B_n$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{1}{n-j+1}=0$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{1}{n-j+2}=0$$

$$\sum_{j=0}^{n}B_j{n \choose j}\frac{j^2}{(n-j+1)(n-j+2)}=B_n$$

-- $$n=2k$$

$$\frac{n!}{B_n}\sum_{j=0}^{n}\frac{(x-2^j)B_{n-j}B_j}{(n-j)!j!}=(x-1)(1-n)$$

--

$$n=2k$$

$$\lim_{n \to \infty}\frac{n!}{B_n}\sum_{j=0}^{n}\frac{(1-2^{1-j})(1-2^{1+j-n})B_{n-j}B_j}{2^j(n-j)!j!}=-\frac{\pi}{2}$$

$$\lim_{n \to \infty}\frac{n!}{B_n}\sum_{j=0}^{n}\frac{(1-2^{1-j})(1-2^{1+j-n})B_{n-j}B_j}{2^j(n-j)!(j-1)!}=-\frac{\pi}{2}$$

$$\lim_{n \to \infty}\frac{n!}{B_n}\sum_{j=0}^{n}\frac{(1-2^{1-j})(1-2^{1+j-n})B_{n-j}B_j}{4^j(n-j)!j!}=-\frac{\pi}{\sqrt{8}}$$

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$$n=2k$$, $$k\ge0$$

$$\frac{n!}{B_n}\sum_{j=0}^{n}\frac{B_{n-j}B_j}{(n-j)!j!}=1-n$$

$$\frac{n!}{B_n}\sum_{j=0}^{n}\frac{(1-2^{1-j})(1-2^{1+j-n})B_{n-j}B_j}{(n-j)!(j-1)!}=-\frac{n(n-1)}{2}$$

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$$n=2k$$, $$k\ge2$$

$$\frac{n!}{B_n}\sum_{j=0}^{n}\frac{B_{n-j}B_j}{(n-j)!(j-1)!}=-\frac{n(n-1)}{2}$$

$$\frac{n!}{B_n}\sum_{j=0}^{n}\frac{2^jB_{n-j}B_j}{(n-j)!j!}=1-n$$

$$\frac{n!}{B_n}\sum_{j=0}^{n}\frac{2^{-j}B_{n-j}B_j}{(n-j)!j!}=\frac{1-n}{2^n}$$

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$$\sum_{k=1}^{n}\frac{{n \choose k-1}B_k}{k}=-\frac{1}{1+n}$$

$$\sum_{k=1}^{n}\frac{(-1)^k{n \choose k-1}B_k}{k}=\frac{n}{1+n}$$

$$\sum_{k=0}^{n}(-1)^k{n \choose k}{k \choose \left\lfloor \frac{k}{2}\right\rfloor}2^{n-k}={2n \choose n}-{2n \choose n+1}=C_n$$

$$\sum_{k=0}^{n}{n \choose k}{k \choose \left\lfloor \frac{k}{2}\right\rfloor}2^{n-k}={2n \choose n}+{2n \choose n+1}$$

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$$\sum_{k=0}^{n}\frac{(-1)^k}{2n+1}{n \choose k}{n+k \choose k}=\frac{1}{2n+1}$$

$$\sum_{k=0}^{n}\frac{(-1)^{k+1}}{2n-1}{n \choose k}{n+k \choose k}=2n+1$$

$$\sum_{k=0}^{2n}\frac{(-1)^k}{2n+1}{2n \choose k}{2n+k \choose k}=\frac{1}{4n+1}$$

$$\sum_{k=0}^{2n}\frac{(-1)^{k+1}}{2n-1}{2n \choose k}{2n+k \choose k}=4n+1$$

$$\sum_{k=0}^{2n}\frac{(-1)^{k+1}}{2n-1}{2n \choose k}{2n+k \choose k}k=2n$$

$$\sum_{k=0}^{2n}\frac{(-1)^{k}}{2n-1}{2n \choose k}{2n+k \choose k}k^2=2n^2$$

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$$\sum_{j=0}^{2(n-1)}(-1)^jj^2{2j+2k \choose j}{2n+2k+j-1\choose 2(n-1)-j}=[2n^2+2n(k-1)-2k]^2$$

$$\sum_{k=0}^{2n}(-1)^k{2n \choose k}{2n+k \choose k}F_k=0$$

$$\sum_{k=j}^{n}{k \choose j}{n \choose k}{n-j \choose k}={2(n-j)\choose n-j}{n-j \choose j}$$

$$\sum_{k=j}^{n}{k \choose j}{n \choose k}{n-j \choose k+j}={2j \choose j}{2(n-j)\choose n-j}{n-j \choose 2j}{n+j \choose j}^{-1}$$

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$$\sum_{k=m}^{n}{k \choose k-m}^2{n \choose k}^2={n \choose m}^2{2(n-m)\choose n-m}$$

$$\sum_{k=m}^{n}{k \choose k-m}^2{n \choose n-k}^2={n \choose n-m}^2{2(n-m)\choose n-m}$$

$$\sum_{k=m}^{n}x^k{k \choose k-m}{n \choose k}=x^m(x+1)^{n-m}{n \choose m}$$

$$\sum_{k=m}^{n}x^k{k \choose k-m}{n \choose n-k}=x^m(x+1)^{n-m}{n \choose n-m}$$

$$\sum_{k=m}^{n}{k \choose k-m}{n \choose k}^2=\frac{\Gamma(m)\Gamma(2n+1-m)}{\Gamma^2(n+1)}{n \choose m}^2$$

$$\sum_{k=m}^{n}{k \choose k-m}{n \choose n-k}^2=\frac{\Gamma(m+1)\Gamma(2n+1-m)}{\Gamma^2(n+1)}{n \choose n-m}^2$$

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$$\sum_{q=0}^{2n}\frac{{2q \choose q}{2n \choose q}q^2}{(-2)^q(2q-1)}=0$$

$$\sum_{q=0}^{2n}\frac{(-2)^q(2q-1)}=-(4n+1)\frac{(2n-1)!!}{(2n)!!}$$

$$\sum_{q=0}^{2n}\frac{(-2)^q}=\frac{(2n-1)!!}{(2n)!!}$$

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$$\sum_{k=0}^{2n}(-1)^k\frac{1}{2^k}{2n+k\choose k}{2n \choose k}=(-1)^n\frac{(2n-1)!!}{(2n)!!}$$

$$\sum_{k=0}^{2n}(-1)^k\frac{k^2}{2^k}{2n+k\choose k}{2n \choose k}=-(-1)^n\frac{(2n+1)!!}{(2n-2)!!}$$ --

$$\sum_{q=0}^{k}q{2k \choose q}{r-2k+1\choose k-q+1}=2k{r \choose k}-k{2k \choose k}$$

$$\sum_{q=0}^{k}q{2k \choose q}{r-2k+1\choose k-q}=2k{r \choose k-1}$$

$$\sum_{q=0}^{k}(-1)^qq{k \choose q}{k-q-n \choose k}=kn$$

$$\sum_{q=0}^{k}\frac{(-1)^q}{(q+1)^2}{k \choose q}{k-q+n \choose k}=\frac{\Gamma(k+n+2)}{\Gamma(k+2)\Gamma(n+2)}\left(H_{n+1}+H_{k+1}-H_{n+k+1}\right)$$

-

$$\sum_{k=1}^{n}\frac{{n \choose k}^2}{(k+1)^2}\left\lfloor a+\frac{k}{n} \right\rfloor=\frac{2a(2n+1){2n\choose n}-(n+1)(an^2+2an+2a-1)}{(n+1)^3}$$

$$\sum_{k=1}^{n} \frac{1}{k}\left\lfloor a-\frac{k}{n} \right\rfloor=(a-1)H_n$$

$$\sum_{k=1}^{n} \frac{k+1}\left\lfloor 2-\frac{k}{n} \right\rfloor=\frac{2^{n+1}-n-2}{n+1}$$

$$\sum_{k=1}^{n} \frac{k+1}\left\lfloor a+\frac{k}{n} \right\rfloor=\frac{a(2^{n+1}-n-2)}{n+1}+\frac{1}{n+1}$$

$$\sum_{k=1}^{n} \frac{{n \choose k}^2}{k+1}\left\lfloor 2-\frac{k}{n} \right\rfloor=\frac{\Gamma(2n+2)}{\Gamma^2(n+2)}-1$$

$$\sum_{k=1}^{n} \frac{{n \choose k}^2}{k+1}\left\lfloor a+\frac{k}{n} \right\rfloor=a\left(\frac{\Gamma(2n+2)}{\Gamma^2(n+2)}-1\right)+\frac{1}{n+1}$$

$$\sum_{k=1}^{n} (-1)^k \frac{(k+1)^2}\left\lfloor 2-\frac{k}{n} \right\rfloor=\frac{H_{n+1}}{n+1}-1$$

-

$$\sum_{j=k}^{n}\frac{(-1)^jj^s}{(j+q)^t}{j \choose k}{n \choose j}=F(s,q,t)$$

$$F(-1,1,1)=(-1)^k\left(\frac{1}{k}-\frac{1}{n+1}\right)$$

$$F(-1,1,2)=(-1)^k\frac{H_k-H_{n+1}}{n+1}+(-1)^k\left(\frac{1}{k}-\frac{1}{n+1}\right)$$

$$F(0,1,2)=(-1)^k\frac{H_k-H_{n+1}}{n+1}$$

$$F(0,q,2)=\sum_{j=k}^{n}\frac{(-1)^j}{(j+q)^2}{j \choose k}{n \choose j}=(-1)^k\left[H_{k+q-1}-H_{n+q}\right]\cdot \frac{\Gamma(n+1)\Gamma(k+q)}{\Gamma(k+1)\Gamma(n+q+1)}$$

-

$$\sum_{j=0}^{2n}\frac{(-1)^j}{j+q}{j \choose k}{n \choose j}=(-1)^k{n \choose k}\operatorname{B}(k+q,n-k+1)$$

$$\sum_{j=0}^{2n}\frac{(-1)^j}{(j+q)^2}{j \choose k}{n \choose j}=(-1)^k{n \choose k}\operatorname{B}(k+q,n-k+1)\left[H_{n+q}-H_{k+q-1}\right]$$

$$\sum_{j=0}^{2n}\frac{(-1)^j}{(j+1)^2}{2n \choose j}^2=\frac{1}{(2n+1)^2}$$

$$\sum_{j=0}^{2n}\frac{(-1)^j}{j+1}{2n \choose j}=\frac{1}{2n+1}$$

$$\sum_{j=0}^{n}\frac{(-1)^j}{(j+1)^2}{n \choose j}=\frac{H_{n+1}}{n+1}$$

-

$$\sum_{n=1}^{\infty}\frac{2^n}{n(n+1){2n \choose n}}\left[n+(n^2-1)H_n\right]=2(\pi-1)+\left(1-\frac{\pi}{2}\right)\left(1+\frac{\pi}{2}\right)$$

$$\sum_{n=0}^{\infty}\frac{2^n H_n}=2G+\frac{\pi}{2}(1-\ln 2)$$

$$\sum_{n=1}^{\infty}\frac{2^n H_n}=2G+\frac{\pi}{2}(2-\ln 2)$$

-

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i+j}}{(i+j)^s}=\eta(s)-\eta(s-1)$$

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{1}{(i+j)^s}=\zeta(s)-\zeta(s+1)$$

---

$$\prod_{j=1}^{\infty}\left(H_j-\frac{\ln j}{j}\right)^{(-1)^j}=\pi?$$

--

$$n\ge 1$$

$$\prod_{k=0}^{\infty}2\left(1-\frac{n^{1/2^k}}{1+n^{1/2^k}}\right)=\frac{4\ln n}{3n^2-5n+4}$$

$$\prod_{k=0}^{\infty}2\left(1-\frac{1}{1+n^{1/2^k}}\right)=f(n)$$

$$f(2n)=2\frac{2n}{2n-1}\cdot \frac{2n}{2n+1}\cdot\ln(2n)$$

$$f(2n-1)=\frac{(2n+1)^2}{2n(n+1)}\ln(2n-1)$$

$$\lim_{n \to \infty}2\prod_{i=1}^{n}(2\ln(2i))^{-1}\prod_{j=1}^{n}\prod_{k=0}^{\infty}2\left(1-\frac{1}{1+(2j)^{1/2^k}}\right)=\pi$$

$$\prod_{j=1}^{\infty}\frac{1}{(2j)^2-1}\prod_{k=0}^{\infty}(2j)^{1/2^k}=\frac{\pi}{2}$$

$$\prod_{j=1}^{\infty}\frac{1}{(j-\frac{1}{2})(j+\frac{1}{2})}\times\frac{\prod_{k=0}^{\infty}(2j)^{1/2^k}}{\sum_{k=0}^{\infty}\frac{1}{2^k}}=\pi$$

$$\lim_{n \to \infty}\left(\frac{2n+1}{\ln(2n+1)!}\right)^{2n+1}\prod_{j=1}^{n}\frac{(j+1)(j+2)}{(2j+3)^2}\prod_{k=0}^{\infty}4\left(1-\frac{1}{1+(2j)^{1/2^k}}\right)\left(1-\frac{1}{1+(2j+1)^{1/2^k}}\right)=\frac{\pi}{2}$$

$$\lim_{n \to \infty}\frac{1}{2^n}\prod_{j=1}^{n}\frac{(2j)^2-1}{H_{2j}-\gamma}\prod_{k=0}^{\infty}\frac{2}{1+(2j)^{1/2^k}}=1$$

---

$$\sum_{m=k}^{2n-k}(-1)^m{m \choose k}{2n-m \choose k}=(-1)^k{n \choose k}$$

$$\sum_{m=k}^{n-k}(-1)^m{m \choose k}{m \choose n}^{-1}{n-m \choose k}=(-1)^k2^{-2k+n-1}\frac{n}{n-k}{n-k \choose k}{k \choose n}^{-1}=(-1)^k2^{-2k+n-1}\frac{B(n-k,k+1)}{B(n-2k+1,k+1)}\cdot \frac{1}{n-k+1}$$

$$\sum_{m=k}^{n-k}(-1)^m{m \choose k}{m \choose n}^{-1}{n-m \choose k}(m-k)=(-1)^k2^{-2k+n-2}(1-n)\cdot \frac{B(n+1,k-n)}{B(k+1,n-2k)}$$

---

$$\sum_{i=0}^{j}(-1)^ii^x{j\choose i}(j^n+i^n)^z=(-1)^jj!\sum_{k=0}^{z}j^{(z-k)n}S^{(j)}_{kn+x}$$

$$\sum_{i=0}^{j}(-1)^ii^x{j\choose i}(j^n-i^n)^z=(-1)^jj!\sum_{k=0}^{z}(-1)^kj^{(z-k)n}S^{(j)}_{kn+x}$$

--

$$j= nz$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}(j^n-i^n)^z=j!$$

$$\sum_{i=0}^{j}(-1)^ii{j\choose i}(j^n-i^n)^z=j!\cdot\frac{j(j+1)}{2}$$

$$j>nz$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}(j^n-i^n)^z=0$$

--

$$j>nz$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}\frac{(j^n-i^n)^z}{i+1}=\frac{(j^n-(-1)^n)^z}{j+1}$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}\frac{(j^n+i^n)^z}{i+1}=\frac{(j^n+(-1)^n)^z}{j+1}$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}\frac{(j^n-i^n)^z}{i+1}i=-\frac{(j^n-(-1)^n)^z}{j+1}$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}\frac{(j^n+i^n)^z}{i+1}i=-\frac{(j^n+(-1)^n)^z}{j+1}$$

$$\sum_{i=0}^{j}x^i{j\choose i}\frac{j-i}{i+1}=\frac{(1+x)^j-1}{x}$$

$$\sum_{i=0}^{j}(-1)^i{j\choose i}\frac{j^n+j^{n+1}-i^n-i^{n+1}}{i+1}=j^{n}$$ -

$$2\sum_{j=0}^{k}\sum_{i=0}^{j}{k \choose j}{k \choose i}=4^k+{2k \choose k}$$

$$\sum_{j=0}^{k}\sum_{i=0}^{j}{k+1 \choose j+1}^2{k \choose i}=2^{k}{2k\choose k}\frac{2k+1}{k+1}=2^{k-1}{2(k+1)\choose k+1}$$

-

$$\sum_{k=0}^{2n}\frac{p^{2k}}{2k+1}(q+p)^{4n-2k}{4n \choose 2k}=\frac{(2p+q)^{4n+1}-q^{4n+1}}{2(4n+1)}$$

--

$$\sum_{k=0}^{n}\frac{1}{k+1}{n \choose k}(2p)^k(q-p)^{n-k}=\frac{(q-p)^n}{2p(n+1)}\left[(p+q)\left(\frac{q+p}{q-p}\right)^n+p-q\right]$$

$$\sum_{k=0}^{n}\frac{1}{k+1}{n \choose k}(-2p)^k(q+p)^{n-k}=\frac{(q+p)^n}{2p(n+1)}\left[(p-q)\left(\frac{q-p}{q+p}\right)^n+p+q\right]$$

-

$$\sum_{k=0}^{n}\frac{1}{k+1}{n \choose k}p^k(q+p)^{n-k}=\frac{(q+p)^n}{p(n+1)}\left[(2p+q)\left(1+\frac{p}{p+q}\right)^n-p-q\right]$$

$$\sum_{k=0}^{n}\frac{(-1)^k}{k+1}{n \choose k}p^k(q-p)^{n-k}=\frac{(q-p)^n}{p(n+1)}\left[(2p-q)\left(1+\frac{q}{p-q}\right)^n-p+q\right]$$

$$\sum_{k=0}^{n}\frac{1}{k+1}{n \choose k}p^k(q-p)^{n-k}=\frac{(q-p)^n}{p(n+1)}\left[q\left(\frac{q}{q-p}\right)^n+p-q\right]$$

$$\sum_{k=0}^{n}\frac{(-1)^k}{k+1}{n \choose k}p^k(q+p)^{n-k}=\frac{(q+p)^n}{p(n+1)}\left[-q\left(\frac{q}{q+p}\right)^n+p+q\right]$$

$$\sum_{k=0}^{n}k{n \choose k}p^k(q-p)^{n-k}=np\cdot q^{n-1}$$

$$\sum_{k=0}^{n}k{n \choose k}p^k(q+p)^{n-k}=np\cdot (2p+q)^{n-1}$$

-

$$\sum_{i=0}^{n}\frac{1}{2i-1}\cdot\frac{1}{2^{2i}}{2i \choose i}=-\frac{(2n-1)!!}{(2n)!!}$$

$$\sum_{i=0}^{n}\frac{i}{2i-1}\cdot\frac{1}{2^{2i}}{2i \choose i}=\frac{(2n-1)!!}{(2n)!!}\cdot n$$

-

$$\sum_{s=0}^{m}(-1)^s{2s \choose s}{m+s\choose m-s}=(-1)^m$$

$$\sum_{s=0}^{m}(-1)^ss{2s \choose s}{m+s\choose m-s}=(-1)^mm(m+1)$$

$$\sum_{s=0}^{m}(-1)^ss^2{2s \choose s}{m+s\choose m-s}=(-1)^m\frac{[m(m+1)]^2}{2}$$

$$\sum_{s=0}^{m}(-1)^s{2s \choose s}{m+s\choose m-s}\frac{1}{(s+1)^2}=\frac{1}{m(m+1)}$$

$$\sum_{s=0}^{m}(-1)^s{2s \choose s}{m+s\choose m-s}\frac{s}{(s+1)^2}=(-1)^m$$

$$\sum_{s=0}^{m}(-1)^s{2s \choose s}{m+s\choose m-s}\frac{s^2}{(s+1)^2}=(-1)^m\cdot\frac{m(m+1)+(-1)^m}{m(m+1)-(-1)^m}$$

$$\sum_{j=1}^{2n}(-1)^jj{4n-j \choose j}2^{2n-j}=(-1)^n4n$$

$$\sum_{j=1}^{2n-1}(-1)^jj{4n-2-j \choose j}2^{2n-1-j}=(-1)^n$$

$$\sum_{j=k}^{2k}(-1)^{j-k}j{j \choose k}{4k-j \choose j}4^{(2k-j)}={4k+1 \choose 2k+1}\frac{4k}{2k+3}$$

$$\sum_{j=2^k}^{2^{k+1}}(-1)^{j-2^k}j^s{j \choose 2^k}{2^{k+1}-j \choose j}4^{(2^{k+1}-j)}=4^{2^k}\cdot2^{ks}$$

$$\sum_{n=1}^{\infty}\frac{\sigma^2_0(c_n)-4}{c^s_n}=\frac{\zeta^4(s)}{\zeta(2s)}-4\zeta(s)+3$$

-

$$\sigma_k(P_a)+\sigma_k(P^2_a)=(P_a^k+1)^2+1$$

$$\sigma_k(P_a)+2\sigma_k(P^2_a)+\sigma_k(P^3_a)=(P_a^k+1)^3+(P_a^k+3)$$

-

General

$$\sum_{j=0}^{n}(-1)^j{n \choose j}\sigma_0^{n-j}(P_a^{j+1})=f(n)$$

$$f(0)=1$$, $$f(1)=1$$, $$f(2)=-6$$, $$f(3)=-22$$, $$f(4)=-2$$, $$f(5)=332$$

$$1+P(s)+\sum_{n=1}^{\infty}\frac{\sigma_0(P_n^2)}{P_n^s}+\sum_{n=1}^{\infty}\frac{\sigma_0^2(c_n)}{c^s_n}=\frac{\zeta^4(s)}{\zeta(2s)}$$

$$\sum_{n=1}^{\infty}\frac{\sigma_0(P_n^2)}{P_n^s}+\sum_{n=1}^{\infty}\frac{\sigma_0^2(c_n)-1}{c^s_n}=\frac{\zeta^4(s)}{\zeta(2s)}-\zeta(s)$$

$$c_n$$; composite number

$$\lim_{n \to \infty}{\frac{\sigma(n)}{n}}=\frac{5}{2}=\frac{\zeta^2(2)}{\zeta(4)}$$

$$a=$$ odd numbers and $$a\ne d$$, $$d\ne0$$

$$\sigma_k(a)\sigma_k(a+d)=\sigma_k[a(a+d)$$

$$\sigma_k(a)\sigma_k(a+d)\sigma_k(a+2d)=\sigma_k[a(a+d)(a+2d)]$$

$$\sigma_k(a)\sigma_k(a+d)\sigma_k(a+2d)\sigma_k(a+3d)=\sigma_k[a(a+d)(a+2d)(a+3d)]$$

-

$$k\ne1$$

$$\sigma(a)\sigma(ka)-\sigma(ka^2)=(k+1)a$$

$$\sigma_n(a)\sigma_n(ka)-\sigma_n(ka^2)=a^n\sigma_n(k)$$

---

General

$$\sum_{j=0}^{n}(-1)^j{n \choose j}\sigma_k(P_a^{j+1})=(-1)^nP^{2k}_a(P^k_a-1)^{n-1}$$

$$\sigma_k(P_a)-\sigma_k(P^2_a)=-P^{2k}_a$$

$$\sigma^2_k(P_a)-\sigma_k(P^2_a)=P^k_a$$

$$\sigma^3_k(P_a)-\sigma_k(P^2_a)=P^k_a\left[(P^k_a+1)^2+1\right]$$

---

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^kk}{n+k}=-\frac{1}$$

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^kk}{(n+k)^2}=-\frac{H^{'}_{2n}}$$

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^kk}{n+k+a}=-B(n+1,n+1+a)$$

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^kk}{(n+k+a)^2}=-[H_{2n+a}-H_{n+a}]B(n+1,n+1+a)$$

---

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^k}{(n+k+\alpha)}=B(n+1,n+\alpha)$$

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^k}{(n+k+\alpha)^2}=\left[H_{2n+\alpha}-H_{n+\alpha-1}\right]B(n+1,n+\alpha)$$

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^k}{(n+k)}=\frac{1}{n{2n \choose n}}$$

$$\sum_{k=0}^{n}{n \choose k}\frac{(-1)^k}{(n+k)^2}=\frac{H_{2n}-H_{n-1}}{n{2n \choose n}}$$

$$\sum_{k=1}^{2n+1}\frac{{2n+1 \choose k}F_k}{\gcd(2^k,2^{2n+1})}=5^n\cdot \frac{L_{2n+1}}{2^{2n+1}}$$

$$\sum_{k=1}^{2n}\frac{{2n\choose k}F_k}{\gcd(2^k,2^{2n})}=5^n\cdot \frac{F_{2n}}{2^{2n}}$$

$$\sum_{k=1}^{n}\frac{(-1)^k{n\choose k}F_k}{\gcd(2^k,2^{n})}=-\frac{F_{2n}}{2^n}$$

$$\sum_{k=1}^{n}\frac{(-2)^k{n\choose k}F_k}{\gcd(2^k,2^{n})}=-F_n$$

$$n\ge0$$

---

$$\sum_{k=1}^{n}\frac{k{k \choose n}}{\gcd(a^k,a^n)}=n\cdot\frac{(a+1)^{n-1}}{a^n}$$

$$\sum_{k=1}^{n}\frac{k^2{k \choose n}}{\gcd(a^k,a^n)}=n(n+2)\cdot\frac{(a+1)^{n-2}}{a^n}$$

$$\sum_{k=1}^{n}\frac{k^3{k \choose n}}{\gcd(a^k,a^n)}=n(n^2+6n+2)\cdot\frac{(a+1)^{n-3}}{a^n}$$

---

$$\sum_{i=0}^{n}\sum_{j=0}^{n}\frac{(2n)!}{(n-i)!(n-j)!}\cdot\frac{(2i+2j)!}{(2i)!(2j)!}\cdot\frac{(-1)^{i+j}}{(i+j)!}=(4n)!!$$

---

If $$N-2\sum_{k=1}^{N}\frac{(-1)^k k}{\gcd(k,N)}=3$$

$$N(1-N)+2\sum_{k=1}^{N}\frac{k}{\gcd(k,N)}=2$$, then N is a prime number

-

$$\sum_{k=1}^{P_{n+1}}\frac{(-1)^kk}{\gcd(k,P_{n+1})}-\sum_{k=1}^{P_{n}}\frac{(-1)^kk}{\gcd(k,P_n)}=\frac{P_{n+1}-P_{n}}{2}$$

$$\sum_{k=1}^{P_{n}}\frac{(-1)^kk^s{P_n \choose k}}{\gcd(k,P_n)}=P^{s-1}_n(P_{n}-1)$$

$$\frac{1}{P^{s-1}_{n+1}}\sum_{k=1}^{P_{n+1}}\frac{(-1)^kk^s}{\gcd(k,P_{n+1})}-\frac{1}{P^{s-1}_{n}}\sum_{k=1}^{P_{n}}\frac{(-1)^kk^s}{\gcd(k,P_n)}=\frac{P_{n+1}-P_{n}}{2}$$

$$\sum_{k=1}^{P_{n}}\frac{(-1)^kk}{\gcd(k,P_n)}=\frac{P_n-3}{2}$$

$$\sum_{k=1}^{P_{n}}\frac{(-1)^kk^2}{\gcd(k,P_n)}=P_n\cdot\frac{P_n-3}{2}$$

-

$$\sum_{k=1}^{2n+1}\frac{(-1)^k}{gcd(k,2n+1)}=-\frac{1}{2n+1}$$

$$\sum_{k=1}^{2n+1}\frac{(-2n)^k}{gcd(k,2n+1)}=-\frac{2n}{2n+1}$$

$$\sum_{k=1}^{2n+1}\frac{(-1)^k{2k+1 \choose k}}{gcd(k,2n+1)}=-\frac{1}{2n+1}$$

$$\sum_{k=1}^{p_n}\frac{(-1)^k{p_n \choose k}k^{s}}{gcd(k,p_n)}=p_n^{s-1}{(p_n-1)}$$

$$\sum_{k=1}^{p_n}\frac{(-1)^k{p_n \choose k}k^{s}}{gcd(k+1,p_n)}=-(p_n-1)^{s+1}$$

$$\sum_{k=1}^{p_n}\frac{(-1)^k{p_n \choose k}k^{s}}{gcd^t(k,p_n)}=p_n^{s-t}{(p^t_n-1)}$$

$$s\ge1$$ and $$p_n$$; a prime number

--

$$\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{i}{(i+j)^2}=\ln{2}$$

$$\sum_{q=\alpha}^{p}(-1)^{q}{p \choose q}{q \choose \alpha}\frac{q^{p-\alpha}}{q-1}=\frac{p}{\alpha(\alpha-1)}$$

$$\sum_{q=\alpha}^{p}(-1)^{q}{p \choose q}{q \choose \alpha}\frac{q^{p-\alpha}}{q-1}=\frac{1}{p+1}$$

--

$$\sum_{k=0}^{n}\frac{k^2(k+1)!}{n^k}{n \choose k}=2n^2$$

--

$$\sum_{m=s}^{2s}\sum_{k=0}^{s}{2s\choose s}{s \choose k}{m \choose k}{k \choose m-s}\frac{m(-1)^m}{(s+1)(4k^2-1)(2m-2k-1)}=\frac{2^{2s-1}}{2s+1}(-1)^{s-1}$$

$$\sum_{m=s}^{2s}\sum_{k=0}^{s}{2s\choose s}{s \choose k}{m \choose k}{k \choose m-s}\frac{k(-1)^m}{(2k-1)(2m-2k-1)}=2^{2s-1}(-1)^{s}$$

$$\sum_{m=s}^{2s}\sum_{k=0}^{s}{2s\choose s}{s \choose k}{m \choose k}{k \choose m-s}\frac{m(-1)^m}{(2k-1)(2m-2k-1)}=2^{2s}(-1)^{s-1}$$

$$\sum_{m=s}^{2s}\sum_{k=0}^{s}{2s\choose s}{s \choose k}{m \choose k}{k \choose m-s}\frac{mk(-1)^m}{(s+1)(2k-1)(2m-2k-1)}=3\cdot2^{2s-2}(-1)^{s-1}$$

$$\sum_{m=s}^{2s}\sum_{k=0}^{s}{2s\choose s}{s \choose k}{m \choose k}{k \choose m-s}\frac{m^2(-1)^m}{(s+1)(2k-1)(2m-2k-1)}=3\cdot2^{2s-1}(-1)^{s-1}$$

---

Pythagoras triples

$$\sum_{k=0}^{2n+(2n+1)}\frac{k}{\cos^2\left(\frac{k\pi}{n}\right)}=(2n+1)^4-(2n)^4$$

$$n\ge0$$

-

$$\sum_{k=0}^{n}k^3{n\choose k}=n(n+1)(n+2)\cdot 2^{n-3}-n\cdot 2^{n-2}$$

-

$$\sum_{k=0}^{3n}(-3)^k{6n \choose2k+1}=0$$

$$\sum_{k=0}^{n}(-3)^k{2n \choose2k+1}=(-1)^n\cdot2^{2n-1}$$

$$n\ne3k$$, $$k=1,2,3,...$$

---

$$\sum_{k=0}^{n}{n \choose k}x^{n-k}=(x+1)^n$$

$$\sum_{k=0}^{n}k{n \choose k}x^{n-k}=n(x+1)^{n-1}$$

$$\sum_{k=0}^{n}k^2{n \choose k}x^{n-k}=n(n-1)(x+1)^{n-2}+n(x+1)^{n-1}$$

$$\sum_{k=0}^{n}k^3{n \choose k}x^{n-k}=n(n-1)(n-2)(x+1)^{n-3}+n^2(x+1)^{n-1}$$

--

$$\sum_{n=1}^{\infty}n\sum_{j=2}^{\infty}\frac{(-1)^j}{j^s}\left(1-\frac{1}{j}\right)^{n-1}=1-\eta(s-2)$$

$$\sum_{n=1}^{\infty}n\sum_{j=2}^{\infty}\frac{(-1)^j}{j^s}\left(1-\frac{1}{j}\right)^{n}=\eta(s-1)-\eta(s-2)$$

$$\sum_{n=1}^{\infty}\sum_{j=2}^{\infty}\frac{(-1)^j}{j^s}\left(1-\frac{1}{j}\right)^{n-1}=1-\eta(s-1)$$

$$\sum_{n=1}^{\infty}\sum_{j=2}^{\infty}\frac{(-1)^j}{j^s}\left(1-\frac{1}{j}\right)^{n}=\eta(s)-\eta(s-1)$$

$$\sum_{n=1}^{\infty}\sum_{j=2}^{\infty}\frac{(-1)^j}{j(j+1)}\left(1-\frac{1}{j}\right)^{n-1}=\eta(1)-\eta(0)$$

---

$$\sum_{k=0}^{n}\left[{2n \choose 2k}^x-{2n \choose 2k-2}^x\right]=1$$

$$\sum_{k=0}^{n}\left[{2n \choose 2k-1}^2-{2n \choose 2k-2}^2\right]=1-(-1)^n {2n \choose n}$$

$$\sum_{k=0}^{n}\left[{2n \choose 2k-2}^2-{2n \choose 2k-3}^2\right]=-1+(2n)^2+(-1)^n {2n \choose n}$$

$$\sum_{k=0}^{n}\sum_{j=0}^{3}(-1)^j{2n \choose 2k-j}^2=-3+(2n)^2+4{2n \choose n}$$

-

$$\sum_{k=0}^{n}\left[{2n \choose 2k}^3-{2n \choose 2k-1}^3\right]=(-1)^n {2n \choose n}{3n \choose n}$$

$$\sum_{k=0}^{n}\left[{2n \choose 2k-1}^3-{2n \choose 2k-2}^3\right]=1-(-1)^n {2n \choose n}{3n \choose n}$$

$$\sum_{k=0}^{n}\left[{2n \choose 2k-2}^3-{2n \choose 2k-3}^3\right]=-1+(2n)^3+(-1)^n {2n \choose n}{3n \choose n}$$

$$\sum_{k=0}^{n}\sum_{j=0}^{3}(-1)^j{2n \choose 2k-j}^3=-3+(2n)^3+4{2n \choose n}{3n \choose n}$$

--

$$\sum_{k=0}^{n}{2n \choose 2k}^3-\sum_{k=0}^{n-1}{2n \choose 2k+1}^3=(-1)^n{2n \choose n}{3n \choose n}$$

$$\sum_{k=0}^{n}(-1)^k\left[{2n \choose 2k}-{2n \choose 2k-1}\right]=(-1)^n 2^n$$

$$\sum_{k=0}^{n}\frac{1}{2k+1}{2n+1 \choose 2k}=\frac{2^{2n}}{n+1}$$

$$\sum_{l=0}^{n}\frac{(-2)^{l}}{2l+1}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n}\frac{(-2)^{k}}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\sum_{l=0}^{n}\frac{(-4)^{l}}{2l+1}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n}\frac{(-4)^{k}}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}$$

--

$$\sum_{l=0}^{n}(-2)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n}\frac{(-2)^{k}}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\frac{1}{2n+1}$$

$$\sum_{l=0}^{n}(-2)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n}(-2)^{k}\frac{k}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\frac{n}{2n+1}$$

$$\sum_{l=0}^{n}(-2)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n}(-2)^{k}\frac{k^2}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\frac{n}{4n^2-1}$$

--

$$\sum_{l=0}^{n}(-4)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{m}(-4)^{k}\frac{k}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=-\frac{2m}{(2m-2n+1)(2m-2n-1)}$$

$$\sum_{l=0}^{n}(-4)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{m}(-4)^{k}\frac{k^2}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\frac{2m(2m+2n-1)}{(2m-2n-3)(2m-2n-1)(2m-2n+1)}$$

--

$$\sum_{l=0}^{n}(-2)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n+1}(-2)^{k}\frac{1}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\frac{2}{2n+3}$$

$$\sum_{l=0}^{n}(-2)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n+1}(-2)^{k}\frac{k}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=\frac{(n+1)(2n-1)}{(2n+1)(2n+3)}$$

$$\sum_{l=0}^{n}(-2)^{l}{n \choose l}{2l \choose l}^{-1}\sum_{k=0}^{n+1}(-2)^{k}\frac{k^2}{2k+1}{n \choose k}{2k \choose k}^{-1}{k+l \choose l}=-\frac{n+1}{2n+3}$$

$$\sum_{k=0}^{2n}(-1)^k{4n \choose 2k}{2n \choose k}^{-1}k=-\frac{n}{2n-1}$$

$$\sum_{k=0}^{2n}(-1)^k{4n \choose 2k}^{-1}{2n \choose k}=\frac{4n+1}{2n+1}$$

$$\sum_{k=0}^{2n}(-1)^k{4n \choose 2k}^{-1}{2n \choose k}k=n\cdot\frac{4n+1}{2n+1}$$

$$\sum_{k=0}^{2n}(-1)^k{4n \choose 2k}{2n \choose k}^{-1}\frac{1}{2k+1}=\frac{1}{4n+1}$$

$$\sum_{k=0}^{2n}(-1)^k{4n \choose 2k}{2n \choose k}^{-1}\frac{k^2}{2k+1}=-\frac{n}{4n+1}$$

$$\prod_{j=1}^{\infty}\left(1+\frac{1}{j^2-j+\frac{1}{4}}\right)=\cosh{\pi}$$

$$\prod_{j=1}^{\infty}\left(1-\frac{1}{\left(j^2-j+\frac{1}{4}\right)^2}\right)=\cosh{\pi}$$

$$\prod_{j=-\infty}^{\infty}\left(1-\frac{1}{\left(j^2-j+\frac{1}{4}\right)^2}\right)=\cosh^2{\pi}$$

---

$$\prod_{j=k}^{\infty}\left(1-\frac{1}{j^2+j+\frac{1}{4}}\right)=\frac{2k-1}{2k+1}$$

$$\prod_{j=-k}^{\infty}\left(1-\frac{1}{j^2+j+\frac{1}{4}}\right)=\frac{2k+1}{2k-1}$$

$$\prod_{j=1}^{\infty}\left(1-\frac{1}{\left(j^2+j+\frac{1}{4}\right)^2}\right)=\frac{\cosh{\pi}}{15}$$

$$\prod_{j=1}^{\infty}\left(1-\frac{1}{\left(j^2+j+\frac{1}{4}\right)^3}\right)=\frac{\sinh^2{\left(\pi\frac{\sqrt{3}}{2}\right)}}{63}$$

$$\prod_{j=0}^{\infty}\left(1-\frac{1}{\left(j^2+j+\frac{1}{4}\right)^2}\right)=-\cosh{\pi}$$

$$\prod_{j=-\infty}^{\infty}\left(1-\frac{1}{\left(j^2+j+\frac{1}{4}\right)^2}\right)=-\cosh^2{\pi}$$

$$\prod_{j=-\infty}^{\infty}\left(1-\frac{1}{\left(j^2+j+\frac{1}{4}\right)^3}\right)=-\sinh^4{\left(\pi\frac{\sqrt{3}}{2}\right)}$$

$$\prod_{j=1}^{\infty}\prod_{k=j}^{\infty}\left(1-\frac{1}{j^2+j+\frac{1}{4}}\right)^{(-1)^j}=\frac{\Gamma^4\left(\frac{1}{4}\right)}{8\pi^2}$$

$$\prod_{j=1}^{\infty}\prod_{k=-j}^{\infty}\left(1-\frac{1}{j^2+j+\frac{1}{4}}\right)^{(-1)^j}=\frac{8\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}$$

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$$\prod_{j=2}^{\infty}\prod_{k=j}^{\infty}\left(1+\frac{1}{k^2-1}\right)^{(-1)^j}=\frac{\pi}{2}$$

$$\prod_{j=2}^{\infty}\left(1+\frac{1}{j^2-1}\right)^{(-1)^{j}}=\frac{\pi^2}{8}$$

$$\prod_{j=1}^{\infty}\left[1-\frac{1}{(2j+1)^2}\right]=\frac{\pi}{4}$$

$$\prod_{j=1}^{\infty}\left[1-\frac{1}{(2j+1)^4}\right]=\frac{\pi}{8}\cosh\left(\frac{\pi}{2}\right)$$

$$2\prod_{j=1}^{\infty}\left[1+\frac{1}{(2j-1)^4}\right]=\cos\left(\frac{\sqrt{\pi}}{2}\right)+\cosh\left(\frac{\sqrt{\pi}}{2}\right)$$

$$4\prod_{j=1}^{\infty}\left[1+\frac{1}{(2j+1)^4}\right]=\cos\left(\frac{\sqrt{\pi}}{2}\right)+\cosh\left(\frac{\sqrt{\pi}}{2}\right)$$

$$\prod_{j=1}^{\infty}\left[1-\frac{1}{(aj^2+b)^2}\right]=\frac{b}{\sqrt{b^2-1}}\sinh\left(\pi\sqrt{\frac{b+1}{a}}\right)\sinh\left(\pi\sqrt{\frac{b-1}{a}}\right)\operatorname{csch}^2\left(\pi\sqrt{\frac{b}{a}}\right)$$

$$\prod_{j=1}^{\infty}\left[1-\frac{1}{(j^2+j+a)^2}\right]=\frac{a}{a^2-1}\cosh\left(\frac{\pi}{2}\sqrt{4a+3}\right)\cosh\left(\frac{\pi}{2}\sqrt{4a-5}\right)\operatorname{sech}^2\left(\frac{\pi}{2}\sqrt{4a-1}\right)$$

$$\prod_{j=1}^{\infty}\left[1-\frac{1}{(j+1)^3}\right]=\frac{1}{6\pi^2}\cosh^2\left(\frac{\sqrt{3}}{2}\pi\right)$$

$$\prod_{j=1}^{\infty}\left[1-\frac{1}{(j+1)^4}\right]=\frac{1}{4\pi}\sinh\left(\pi\right)$$

$$\prod_{j=1}^{\infty}\prod_{k=j}^{\infty}\left(1-\frac{1}{(j+1)^2}\right)^{(-1)^{j}}=\frac{\pi}{2}$$

$$\prod_{j=1}^{\infty}\left(1-\frac{1}{(j+1)^2}\right)^{(-1)^{j}}=\frac{\pi^2}{8}$$

--

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n+1} \left\lfloor \frac{\ln n}{\ln 2} \right\rfloor=1-\gamma$$

$$\lim_{j \to \infty}\sum_{n=j}^{\infty}\left(\frac{1}{j\left\lceil\frac{n}{j}\right\rceil}-\frac{1}{n}\right)=\gamma-1$$

$$\lim_{j \to \infty}\sum_{n=j}^{\infty}\left[\frac{1}{j}\left(\frac{1}{\left\lceil\frac{n}{j}\right\rceil}+\frac{1}{\left\lfloor\frac{n}{j}\right\rfloor}\right)-\frac{2}{n}\right]=2\gamma-1$$

--

$$\sum_{j=1}^{2k}\frac{(-1)^j}{j}\left[{2k \choose j}^2-{2k \choose j}\right]=\sum_{j=1}^{k}\frac{(-1)^j}{j}{2k \choose j}$$

$$\sum_{j=1}^{2k}\frac{(-1)^j}{j}{2k \choose j}^2+\sum_{j=1}^{k}\frac{(-1)^j}{j}{2k \choose j}=H_{2k}$$

$$\sum_{j=1}^{2k}\frac{(-1)^j}{j}{2k \choose j}^2+\sum_{j=1}^{k}\frac{(-1)^j}{j}\left[{2k \choose j}+{k \choose j}\right]=H^{*}_{2k}$$

--

$$\sum_{n=1}^{\infty}\frac{H_{n}^2-H_{2n}^2-\ln2[2\gamma+\ln(2n^2)]}{n}=\eta(3)-\eta(1)[\eta(2)-\gamma^2-2\gamma_1]$$

$$\sum_{n=1}^{\infty}\frac{H^{*2}_{2n}-2H_{2n}H^{*}_{2n}-\ln2[2\gamma+\ln(2n^2)]}{n}=\eta(3)-\eta(1)[\eta(2)-\gamma^2-2\gamma_1]$$

---

$$2\sum_{n=1}^{\infty}\frac{H_{2n}^2-[\gamma+\ln(2n)]^2}{2n}=\frac{11}{12}\zeta(3)-\frac{2}{3}\gamma^3-2\gamma\gamma_1-\gamma_2+\eta(1)[\eta(2)-\gamma^2-2\gamma_1]$$

$$2\sum_{n=1}^{\infty}\frac{H_{2n-1}^2-[\gamma+\ln(2n-1)]^2}{2n-1}=\frac{29}{12}\zeta(3)-\frac{2}{3}\gamma^3-2\gamma\gamma_1-\gamma_2-\eta(1)[\eta(2)-\gamma^2-2\gamma_1]$$

$$\sum_{n=1}^{\infty}(-1)^n\frac{H_n^2-(\gamma+\ln n)^2}{n}=-\eta(3)+\eta(1)[\eta(2)-\gamma^2-2\gamma_1]$$

-

$$\sum_{n=1}^{\infty}\frac{H_{2n}+H^{*}_{2n}-\gamma-\ln(4n)}{n}=2\eta^2(1)-\eta(2)-\gamma_1-\frac{1}{2}\gamma^2$$

$$\sum_{n=1}^{\infty}\frac{H_{4n}+H^{*}_{4n}-\gamma-\ln(8n)}{n}=\frac{9\eta^2(1)-5\eta(2)}{2}-\gamma_1-\frac{1}{2}\gamma^2$$

$$\sum_{n=1}^{\infty}(-1)^n\frac{H_n}{n}=\frac{1}{2}\eta^2(1)-\eta(2)$$

$$\sum_{n=1}^{\infty}\frac{H_{2n}-\gamma-\ln(2n)}{2n}=\frac{\eta^2(1)-\gamma_1-\frac{1}{2}\gamma^2}{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{2n-1}-\gamma-\ln(2n-1)}{2n-1}=\frac{\zeta(2)-\eta^2(1)-\gamma_1-\frac{1}{2}\gamma^2}{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}+H_{4n}-2[\gamma+\ln(2n)]}{n}=\frac{11\eta^2(1)-\eta(2)}{4}-2\gamma_1-\gamma^2$$

$$\sum_{n=1}^{\infty}\frac{H_{4n}-\gamma-\ln(4n)]}{n}=\frac{11\eta^2(1)-5\eta(2)}{4}-\gamma_1-\frac{1}{2}\gamma^2$$

$$\sum_{n=1}^{\infty}\frac{H_{4n}-H_{2n}-\ln{2}}{n}=\frac{7\eta^2(1)-5\eta(2)}{4}$$

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$$\sum_{n=1}^{\infty}(-1)^n\cdot\frac{H_{4n}-2H_{2n}-\ln{n}}{n}=\frac{1}{2}\ln^2(1+\sqrt{2})+\frac{1}{8}\eta^2(1)-\gamma\eta(1)+\frac{3}{8}\eta(2)$$

$$\sum_{n=1}^{\infty}(-1)^n\cdot\frac{H_{4n}-2H^{'}_{2n}+\ln{n}}{n}=\frac{1}{2}\ln^2(1+\sqrt{2})+\frac{1}{8}\eta^2(1)+\gamma\eta(1)-\frac{13}{8}\eta(2)$$

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$$\sum_{n=1}^{\infty}(-1)^n\cdot\frac{H_n+\gamma+\ln{n}+\frac{1}{2n}}{n}=-\frac{\pi^2}{8}$$

$$\sum_{n=1}^{\infty}(-1)^n\cdot\frac{H_n+\ln{n}}{n}=\gamma\eta(1)-\eta(2)$$

$$\sum_{n=1}^{\infty}(-1)^n\cdot \left(H_n-\gamma-\ln{n}\right)=\frac{\gamma-\ln{\pi}}{2}$$

-

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n+1}\left(\frac{\ln(n+1)}{\ln2}-\frac{1}{2}\right)=\frac{1}{2}-\gamma$$

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n+2}\left(\frac{\ln(n+2)}{\ln2}-\frac{1}{2}\right)=-\frac{3}{4}-\gamma$$

$$\sum_{n=1}^{\infty}(-1)^n\frac{\ln^2{n}}{n}=\left(2\gamma_1-\frac{\ln^2{2}}{3}+\gamma\ln{2}\right)\ln{2}$$

$$\frac{4}{\pi}\cdot\sum_{n=1}^{\infty}(-1)^n\frac{\ln{(2n-1)}}{2n-1}=\gamma+\ln\frac{4\pi^3}{\Gamma^4\left(\frac{1}{4}\right)}$$

-

$$\sum_{i=1}^{\infty}H_i\sum_{j=0}^{k}(-1)^j{k \choose j}\frac{1}{i+j}=\zeta(2)-\sum_{n=1}^{k-1}\frac{1}{n^2}$$

$$\lim_{n \to \infty}\frac{1}{2}\left(\ln n+\gamma\right)^2-\sum_{k=1}^{n}\frac{H_k}{k+1}=\eta(2)$$

$$\lim_{n \to \infty}\frac{1}{2}\left(\ln n+\gamma\right)^2-\sum_{k=1}^{n}\frac{H_k}{k+2}=1+\eta(2)$$

$$\lim_{n \to \infty}\sqrt{\sum_{k=1}^{n}H_k\left(\frac{1}{k}+\frac{1}{k+1}\right)}-\ln n=\gamma$$

$$\lim_{n \to \infty}\sqrt{1+\sum_{k=1}^{n}H_k\left(\frac{1}{k}+\frac{1}{k+2}\right)}-\ln n=\gamma$$

$$\lim_{n \to \infty}\frac{\left[1-\Gamma\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)\right]\left[1-\Gamma\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)\right]}{\left[2-\Gamma\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)\right]\left[\Gamma\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)\right]}=\frac{\gamma(\gamma-1)}{2\gamma-1}$$

$$\lim_{n \to \infty}n-n\Gamma\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)=\gamma$$

$$\lim_{n \to \infty}n-n\Gamma\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)=1-\gamma$$

$$\lim_{n \to \infty}\frac{1-\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)}{2-\Gamma^y\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^y\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=\frac{x}{y}\cdot \gamma$$

$$\lim_{n \to \infty}\frac{1-\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}{2-\Gamma^y\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^y\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=(1-\gamma)\cdot \frac{x}{y}$$

$$\lim_{n \to \infty}\frac{\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}{2-\Gamma^y\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^y\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=(1-2\gamma)\cdot \frac{x}{y}$$

$$\lim_{n \to \infty}\frac{1-\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)}{1-\Gamma^y\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=\frac{x}{y}\cdot \frac{\gamma}{1-\gamma}$$

$$\lim_{n \to \infty}\frac{1-\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)}{\Gamma^y\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^y\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=\frac{x}{y}\cdot \frac{\gamma}{1-2\gamma}$$

$$\lim_{n \to \infty}\frac{1-\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}{\Gamma^y\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^y\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=\frac{x}{y}\cdot \frac{1-\gamma}{1-2\gamma}$$

$$\alpha\ge2$$

$$\lim_{n \to \infty}n\left(1-\frac{\alpha-\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)}{\alpha-\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}\right)=\frac{1-2\gamma}{\alpha-1}x$$

$$\lim_{n \to \infty}n\left(1-\frac{\alpha+\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)}{\alpha+\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}\right)=\frac{2\gamma-1}{\alpha+1}x$$

--

$$\lim_{n \to \infty}n\times\frac{\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}{\alpha^2-\Gamma^{2x}\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=-x\cdot\frac{2\gamma-1}{\alpha^2-1}$$

$$\lim_{n \to \infty}n\times\frac{\Gamma^x\left(1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-\Gamma^x\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}{\alpha+\beta\Gamma^{y}\left(1+\frac{1}{1+\frac{1}{\Gamma\left(\frac{1}{n}\right)}}\right)}=-x\cdot\frac{2\gamma-1}{\alpha+\beta}$$

$$\lim_{n \to \infty}\zeta\left(\zeta(n)-k\zeta\left(\frac{1}{n}\right)\right)=\zeta\left(1+\frac{k}{2}\right)$$

$$\lim_{n,k \to \infty}\zeta\left(\zeta(n)-\frac{1}{k}\zeta\left(\frac{1}{n}\right)\right)-2k=\gamma$$

$$\lim_{n \to \infty}\frac{x\Gamma^k\left(1+\frac{1}{n}\right)}{\Gamma^x\left(1+\frac{1}{n}\right)-\Gamma^x\left(2+\frac{1}{n}\right)}+n=(k-x)\gamma+\frac{x-1}{2}$$

$$\lim_{n \to \infty}\frac{x\Gamma^k\left(2+\frac{1}{n}\right)}{\Gamma^x\left(1+\frac{1}{n}\right)-\Gamma^x\left(2+\frac{1}{n}\right)}+n=(k-x)\gamma+\frac{x-2k-1}{2}$$

$$\lim_{n \to \infty}\frac{x}{\Gamma^x\left(1+\frac{1}{n}\right)-\Gamma^x\left(2+\frac{1}{n}\right)}+n=-\gamma+\frac{x-1}{2}$$

$$\lim_{n \to \infty}\frac{\Gamma^k\left(1+\frac{1}{n}\right)-1}{\Gamma^x\left(1+\frac{1}{n}\right)-\Gamma^x\left(2+\frac{1}{n}\right)}=\frac{k}{x}\cdot\gamma$$

$$\lim_{n \to \infty}\frac{\Gamma^k\left(2+\frac{1}{n}\right)-1}{\Gamma^x\left(1+\frac{1}{n}\right)-\Gamma^x\left(2+\frac{1}{n}\right)}=\frac{k}{x}\cdot(\gamma-1)$$

--

$$\lim_{n \to \infty}\frac{\Gamma^x\left(1+\frac{1}{n}\right)}{\Gamma\left(1+\frac{1}{n}\right)-\Gamma\left(2+\frac{1}{n}\right)}+n=(x-1)\gamma$$

$$\lim_{n \to \infty}\frac{\Gamma^x\left(2+\frac{1}{n}\right)}{\Gamma\left(1+\frac{1}{n}\right)-\Gamma\left(2+\frac{1}{n}\right)}+n=(1-x)(1-\gamma)-1$$

$$\lim_{n \to \infty}\frac{1}{\Gamma\left(1+\frac{1}{n}\right)-\Gamma\left(2+\frac{1}{n}\right)}+n=-\gamma$$

$$\lim_{n \to \infty}\frac{\Gamma^x\left(1+\frac{1}{n}\right)-1}{\Gamma\left(1+\frac{1}{n}\right)-\Gamma\left(2+\frac{1}{n}\right)}=x\gamma$$

$$\lim_{n \to \infty}\frac{\Gamma^x\left(2+\frac{1}{n}\right)-1}{\Gamma\left(1+\frac{1}{n}\right)-\Gamma\left(2+\frac{1}{n}\right)}=x(\gamma-1)$$

$$\lim_{n \to \infty}\frac{\Gamma\left(1+\frac{1}{n}\right)-1}{\Gamma\left(2+\frac{1}{n}\right)-1}=\frac{\gamma}{\gamma-1}$$

-

$$\lim_{n \to \infty}\Gamma\left(\frac{1}{n}\right)-n=-\gamma$$

$$\lim_{n \to \infty}\Gamma\left(\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)-n=-2\gamma$$

$$\lim_{n \to \infty}\Gamma\left(\frac{1}{\Gamma\left(\frac{1}{\Gamma\left(\frac{1}{n}\right)}\right)}\right)-n=-3\gamma$$

and so on ....

--

$$\lim_{n \to \infty}\Gamma\left(\frac{a}{n}\right)-\Gamma\left(\frac{a}{\Gamma\left(\frac{1}{n}\right)}\right)=\frac{\gamma}{a}$$

$$\lim_{n \to \infty}n\left(1-\Gamma^{a}\left(1+\frac{b}{n}\right)\right)=ab\gamma$$

$$\lim_{n \to \infty}n\left(1-\Gamma^{a}\left(2+\frac{b}{n}\right)\right)=ab(\gamma-1)$$

-

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{n}\right)-\zeta\left(1+\frac{\alpha}{\zeta\left(1+\frac{1}{n}\right)}\right)=-\frac{\gamma}{\alpha}$$

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{n}\right)+\zeta\left(1+\frac{\alpha}{\zeta\left(1-\frac{1}{n}\right)}\right)=\frac{2\alpha+1}{\alpha}\gamma$$

-

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{n}\right)-\frac{n}{\alpha}=\gamma$$

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{\zeta\left(1+\frac{1}{n}\right)}\right)-\frac{n}{\alpha}=\frac{\alpha+1}{\alpha}\gamma$$

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{\zeta\left(1+\frac{1}{\zeta\left(1+\frac{1}{n}\right)}\right)}\right)-\frac{n}{\alpha}=\frac{\alpha+2}{\alpha}\gamma$$

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{\zeta\left(1+\frac{1}{\zeta\left(1+\frac{1}{\zeta\left(1+\frac{1}{n}\right)}\right)}\right)}\right)-\frac{n}{\alpha}=\frac{\alpha+3}{\alpha}\gamma$$

and so on ....

general

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{\zeta\left(1+\frac{\alpha_1}{\zeta\left(1+\frac{\alpha_2}{\zeta\left(1+\frac{\alpha_3}{n}\right)}\right)}\right)}\right)-\frac{n}{\alpha \alpha_1 \alpha_2 \alpha_3}=\frac{\alpha+3}{\alpha}\gamma$$

--

$$\lim_{n \to \infty}\prod_{i=1}^{\infty}\left(\frac{\zeta\left(1+\frac{(i+1)^2}{n}\right)}{\zeta\left(1+\frac{i^2}{n}\right)}\right)^{(-1)^i\zeta(1/n)}=\frac{2}{\pi}$$

$$\lim_{n \to \infty}\prod_{i=1}^{\infty}\left(\frac{\zeta\left(1+\frac{i+1}{n}\right)}{\zeta\left(1+\frac{i}{n}\right)}\right)^{2(-1)^i\zeta(1/n)}=\frac{2}{\pi}$$

$$\lim_{n \to \infty}\frac{\zeta\left(1+\frac{\alpha^2}{n}\right)}{\zeta\left(1+\frac{\alpha}{n}\right)}=\alpha^{\frac{1}{2\zeta(1/n)}}$$

$$\lim_{n \to \infty}\prod_{i=1}^{\infty}\left[\zeta\left(1+\frac{P_i^{2/(P_i^2-1)^2}}{n}\right)\right]=\left(\frac{A^{6}}{\sqrt{2\pi e^{\gamma}}}\right)^{1/\zeta(1/n)}\cdot\prod_{i=1}^{\infty}\left[\zeta\left(1+\frac{P_i^{1/(P_i^2-1)}}{n}\right)\right]$$

$$\lim_{n \to \infty}\prod_{i=1}^{\infty}\left(\frac{\zeta\left(1+\frac{(i+1)^2}{n}\right)}{\zeta\left(1+\frac{i+1}{n}\right)}\right)^{(-1)^i}=\sqrt{\left(\frac{2}{\pi}\right)^{1/\zeta(1/n)}}\cdot \lim_{n \to \infty}\prod_{i=1}^{\infty}\left(\frac{\zeta\left(1+\frac{i^2}{n}\right)}{\zeta\left(1+\frac{i}{n}\right)}\right)^{(-1)^i}$$

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$$\lim_{n \to \infty}\sqrt{n}\zeta\left(1+\frac{\alpha}{n}\right)^{\zeta(1/n)}=\sqrt{\alpha}$$

$$\lim_{n \to \infty}\zeta\left(1+\frac{\alpha}{n}\right)-\frac{n}{\alpha}=\gamma$$

$$\lim_{n \to \infty}\sqrt{n}-\zeta\left(1+\frac{1}{n}\right)^{\zeta(1/n)+1}=0$$

$$\lim_{n \to \infty}\zeta\left(1+\frac{1}{n}\right)\left[1+\zeta\left(1+\frac{1}{n}\right)^{\zeta(1/n)}\right]-n-\sqrt{n}=\gamma$$

$$\lim_{n \to \infty}\left[\zeta\left(\frac{1}{n}\right)\right]^{\zeta(n)}=-\frac{1}{2}$$

-

$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\left((\alpha+1)\left\lceil\frac{n}{k}\right\rceil-\alpha\left\lfloor\frac{n}{k}\right\rfloor-\frac{n}{k}\right)=\gamma+\alpha$$

$$\sum_{k=0}^{n}(n-k){n \choose k}F_{k}=nF_{2(n-1)}$$

$$\sum_{k=0}^{n}(-1)^k(n-k){n \choose k}F_{k}=-nF_{n-1}$$

$$\sum_{k=0}^{n}(-1)^k(n-k){n \choose k}F_{2k}=(-1)^nnF_{n-1}$$

-

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j}{a(j+1)}=a\ln\left(\frac{a}{a-1}\right)-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j^2}{(j+1)(j+2)}=\frac{\pi^2-9}{9}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{2j^2}{(j+1)(2j+1)}=\frac{\pi^2-4}{4}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{4j^2}{(j+1)(4j+1)}=2G+\frac{\pi^2-4}{4}$$

$$\lim_{n \to \infty}\sum_{k=2}^{2n}(-1)^k\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j}{k(j+1)}=\ln\left(\frac{\pi^{3/2}2^{1/6}}{A^6}\right)$$

--

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{1}{2})}{(j+\frac{1}{2})^2}=2G-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\left(\frac{j-\frac{1}{2}}{j+\frac{1}{2}}\right)^s=\lambda(s)-1$$

-

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{1}{3})}{(j+1)(j+\frac{1}{3})}=\frac{\pi}{\sqrt{3}}-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{2}{3})}{(j+1)(j+\frac{2}{3})}=2-\frac{\pi}{\sqrt{3}}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{1}{4})}{(j+1)(j+\frac{1}{4})}=\frac{\pi}{2}-1+\ln{2}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{3}{4})}{(j+1)(j+\frac{3}{4})}=1-\frac{\pi}{2}+\ln{2}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{1}{6})}{(j+1)(j+\frac{1}{6})}=\frac{\pi}{\sqrt{3}}-1+2\ln{2}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{2k-1}{2})}{(j+1)(j+\frac{2k-1}{2})}=\ln{4}-1-H_{(2k-2)}^{*}$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{(j-\frac{1}{2})^2}{j(j+\frac{1}{2})}=\frac{\pi}{2}-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j}{2j+1}=\frac{\pi}{2}-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{(j-\frac{1}{2})}{j(j+\frac{1}{2})^2}=\frac{\operatorname{shi(2)}}{2}-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{(j-\frac{1}{2})}{j(j+\frac{1}{2})}=\frac{\sqrt{\pi}\operatorname{erfi(1)}}{2}-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{(j-\frac{1}{2})}{(j+1)(j+\frac{1}{2})}=\sqrt{\pi}\operatorname{erfi(1)}-e$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-\frac{1}{2})}{(j+1)(j+\frac{1}{2})}=\ln{4}-1$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j^2(j-\frac{1}{2})}{(j+1)^2(j+\frac{1}{2})}=\ln{16}-1-\zeta(2)$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j^3(j-\frac{1}{2})}{(j+1)^3(j+\frac{1}{2})}=\ln{16^2}-1-2\zeta(2)-\zeta(3)$$

$$\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{(j-\frac{1}{2})^2}{(j+1)(j+\frac{1}{2})}=\pi-3$$

-

$$\sum_{j=0}^{n}(-1)^j{n \choose j}j(n-j)^{n-1}=-n!$$

$$\sum_{j=0}^{n}(-1)^j{n \choose j}j(n-\alpha j)^{n-1}=-n!\alpha^{n-1}$$

$$\sum_{j=0}^{n}(-1)^j{n \choose j}j^2(n-j)^{n-1}=n!\cdot \frac{n^2-3n}{2}$$

$$\sum_{j=0}^{n}(-1)^j{n \choose j}j^2(n-2j)^{n-1}=n!\cdot 2^{n-1}\cdot n$$

$$\prod_{k=1}^{4n}\left[\sin\left(\alpha+\frac{k\pi}{4n}\right)+\cos\left(\alpha+\frac{k\pi}{4n}\right)\right]=(-1)^{n-1}\frac{\sin(4n \alpha)}{2^{2n-1}}$$

$$\prod_{k=1}^{4n-2}\left[\sin\left(\alpha+\frac{k\pi}{4n-2}\right)+\cos\left(\alpha+\frac{k\pi}{4n-2}\right)\right]=(-1)^{n-1}\frac{\cos[(4n-2)\alpha]}{2^{2n-2}}$$

$$\prod_{k=1}^{4n}\left[\sin\left(\alpha+\frac{k\pi}{4n}\right)+\cos\left(\alpha+\frac{k\pi}{4n}\right)\right]^{(-1)^k}=(-1)^{n-1}\left[\tan(2n\alpha)\right]^{(-1)^n}$$

$$\prod_{k=1}^{4n-2}\left[\sin\left(\alpha+\frac{k\pi}{4n-2}\right)+\cos\left(\alpha+\frac{k\pi}{4n-2}\right)\right]^{(-1)^n}=(-1)^n\left[\frac{1-\tan[(2n-1)\alpha]}{1+\tan[(2n-1)\alpha]}\right]^{(-1)^n}$$

$$\sum_{k=0}^{2n}(-1)^k{2k \choose k}{4n-2k \choose 2n-k}k=n{2n \choose n}\cdot2^{2n}$$

$$\sum_{k=0}^{2n}(-1)^k{2k \choose k}{4n-2k \choose 2n-k}\frac{1}{1-2k}={2n \choose n}\cdot2^{2n}$$

$$\sum_{k=0}^{2n}(-1)^k{2k \choose k}{4n-2k \choose 2n-k}\frac{k^2}{2k-1}=\frac{n}{2}\cdot{2n \choose n}\cdot2^{2n}$$

--

$$\sum_{r=0}^{3n-2}(-1)^r{2(3n-2)-r\choose r}\frac{r^2}{2(3n-2)-r}=3-4n$$

$$\sum_{r=0}^{3n-1}(-1)^r{2(3n-1)-r\choose r}\frac{r^2}{2(3n-1)-r}=2n-1$$

$$\sum_{r=0}^{3n-1}(-1)^r{2(3n-1)-r\choose r}\frac{r^3}{2(3n-1)-r}=3(2n-1)^2$$

$$\sum_{r=0}^{3n}(-1)^r{6n-r\choose r}\frac{r^2}{6n-r}=2n$$

$$\sum_{r=0}^{3n}(-1)^r{6n-r\choose r}\frac{r^3}{6n-r}=2n$$

---

$$\sum_{i=1}^{\infty}(-1)^i\left(H_{i,s}-\zeta(s)\right)=\frac{\zeta(s)}{2^s}$$

$$\sum_{j=2}^{\infty}\sum_{i=1}^{\infty}(-1)^i\left(H_{i,j}-\zeta(j)\right)=\ln{2}$$

$$\sum_{j=1}^{\infty}\frac{1}{2j+1}\sum_{i=1}^{\infty}(-1)^i\left(H_{i,2j}-\zeta(2j)\right)=\frac{1-\ln{2}}{2}$$

$$\sum_{j=1}^{\infty}\frac{1}{j}\sum_{i=1}^{\infty}(-1)^i\left(H_{i,2j}-\zeta(2j)\right)=\ln\frac{\pi}{2}$$

---

$$\sum_{i=0}^{n}{n \choose i}2^i(n-i)!i!=n!\left(2^{n+1}-1\right)$$

$$\sum_{i=0}^{n}{n \choose i}^22^i(n-i)!i!=3^nn!$$

$$\sum_{i=0}^{n}{n \choose i}^2 {\alpha}^i(n-i)!i!={(\alpha+1)}^nn!$$

$$\prod_{r=1}^{2^k}\Gamma\left(\frac{r+2\alpha}{2^k}\right)^{(-1)^r}=\frac{(2\alpha)!!}{(2\alpha-1)!!}\cdot\frac{1}{\sqrt{2^{k-1}\pi}}$$

$$\prod_{r=1}^{2^k}\Gamma\left(\frac{r+2\alpha-1}{2^k}\right)^{(-1)^r}=\frac{(2\alpha-1)!!}{(2\alpha-2)!!}\cdot\frac{\sqrt{2^{k-1}\pi}}{2^{k}}$$

-

$$\prod_{r=1}^{2k}\Gamma\left(\frac{2r-1}{k}\right)^{(-1)^r}=f(k)$$, are rational values for $$k\ne2^n,$$ for $$n=1,2,3,...$$

$$\prod_{r=1}^{2^{k+1}}\Gamma\left(\frac{r}{2^k}\right)^{(-1)^r}=\frac{(2^k-2)!!}{(2^k-1)!!}\cdot \frac{2}{\pi}$$

$$\prod_{r=1}^{2^{k}}\Gamma\left(\frac{r}{2^k}\right)^{(-1)^r}=\frac{1}{\sqrt{2^{k-1}\pi}}$$ $$\prod_{r=1}^{2^k}\Gamma\left(1+\frac{r}{2^k}\right)^{(-1)^r}=\frac{(2^k-2)!!}{(2^k-1)!!}\cdot \frac{2^k}{\sqrt{2^{k-1}\pi}}$$

$$\prod_{r=1}^{2^k-1}\Gamma\left(1-\frac{r}{2^k}\right)^{(-1)^r}=\frac{1}{\sqrt{2^{k-1}\pi}}$$

$$\prod_{r=1}^{2^k}\Gamma\left(\alpha+\frac{r}{2^k}\right)^{(-1)^r}=f(\alpha)$$

$$\prod_{r=1}^{2^k}\Gamma\left(\alpha-\frac{r}{2^k}\right)^{(-1)^r}=g(\alpha)$$

---

$$\prod_{r=1}^{6}\Gamma\left(\frac{r}{3}\right)^{(-1)^r{5 \choose r-1}}=\frac{\Gamma^{9}\left(\frac{1}{3}\right)}{6^5}$$

$$\prod_{r=1}^{5}\Gamma\left(\frac{2r-1}{5}\right)^{(-1)^r}=\frac{1}{2\phi}$$

$$\prod_{r=1}^{4k-2}\Gamma\left(\frac{2r-1}{4k-2}\right)^{(-1)^r}=f(k)$$

-

$$\prod_{r=1}^{2k+1}\Gamma\left(\frac{r}{2k+1}\right)^{(-1)^r}=\frac{(2k-1)!!}{(2k)!!}$$

$$\prod_{r=1}^{2kn-1}\Gamma\left(\frac{r}{k}\right)^{(-1)^r}=F(n,k)$$

$$\prod_{r=1}^{k}\Gamma\left(\frac{r^2}{k}\right)^{(-1)^r}=F(n,k)$$

$$\prod_{r=1}^{k-1}\Gamma\left(\frac{r^2}{k}\right)^{(-1)^r}=F(n,k)$$

$$k=1,3,5,...$$

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$$\prod_{r=1}^{n}\Gamma\left(\frac{2r}{4n-3}\right)^{(-1)^r}=\frac{(2n-3)!!}{(2n-2)!!}$$

$$\prod_{r=1}^{n}\Gamma\left(\frac{2r}{4n-1}\right)^{(-1)^r}=\frac{(2n-2)!!}{(2n-1)!!}\cdot \frac{2}{\pi}$$

$$\prod_{r=1}^{4}\Gamma\left(\frac{3r}{5}\right)^{(-1)^r}=\frac{2^{-8/5}}{7+7^{-1}}\cdot\frac{\Gamma^2\left(\frac{1}{10}\right)}{\phi\pi}$$

$$\prod_{r=1}^{8}\Gamma\left(\frac{r}{3}\right)^{(-1)^r}=\frac{\Gamma\left(\frac{8}{3}\right)}{2\Gamma\left(\frac{7}{3}\right)}$$

$$\prod_{r=1}^{14}\Gamma\left(\frac{r}{5}\right)^{(-1)^r}=\frac{21\phi}{\pi}\cdot\frac{\Gamma^2\left(\frac{7}{10}\right)}{2^{26/5}}$$

-

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{1}{2j+1}=\frac{(-1)^n}{2n+1}$$

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{1}{2j-1}=-(-1)^n(2n+1)$$

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{1}{2j+3}=-\frac{(-1)^n}{(2n-1)(2n+1)(2n+3)}$$

$$\sum_{j=0}^{n\ge2}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{1}{(j+2)^2}=-(-1)^n\frac{(n+2)!}{(n-2)!}$$

$$n\ge \alpha$$

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{1}{(j+\alpha)^2}=-\frac{(-1)^{n+\alpha}}{(\alpha-1)!^2}\cdot\frac{(n+\alpha)!}{(n-\alpha)!}$$

-

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{j}{(j+1)^2}=-\frac{(-1)^n}{n(n+1)}$$

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{j^2}{(j+1)^2}=1+\frac{(-1)^n}{n(n+1)}$$

$$\sum_{j=0}^{n}(-1)^{n+j}{n \choose j}{n+j \choose j}\frac{j^3}{(j+1)^2}=\frac{[n(n+1)-1]^2+2[1+(-1)^n]}{n(n+1)}$$

--

$$\sum_{n=2}^{\infty}(-1)^n\frac{\lceil \log_2(n)\rceil }{n-1}=\ln{2}-\gamma$$

$$\sum_{n=2}^{\infty}(-1)^n\frac{\lceil \log_2(n)\rceil }{n}=\gamma-\ln{2}$$

$$\sum_{n=2}^{\infty}(-1)^n\frac{\lceil \log_2(n)\rceil }{n+1}=1-2\gamma$$

$$\sum_{n=2}^{\infty}(-1)^n\frac{\lceil \log_2(n)\rceil }{n+2}=1-2\gamma$$

$$\sum_{n=2}^{\infty}(-1)^n\frac{\lceil \log_2(n)\rceil }{n+3}=1-\ln{\pi}$$

---

$$\sum_{n=0}^{\infty}\frac{1}{16^{n+1}}\left(\frac{64}{8n+1}+\frac{16}{8n+2}-\frac{64}{8n+4}-\frac{16}{8n+5}-\frac{16}{8n+6}-\frac{1}{8n+10}+\frac{2}{8n+12}\right)=\pi$$

$$\sum_{n=0}^{\infty}\frac{1}{16^{n+1}}\left(\frac{64}{8n+1}-\frac{64}{8n+4}-\frac{16}{8n+5}+\frac{32}{8n+6}+\frac{2}{8n+12}-\frac{3}{8n+14}\right)=\pi$$

$$\sum_{n=0}^{\infty}\frac{1}{16^{n+1}}\left(\frac{832}{8n+1}-\frac{192}{8n+2}-\frac{832}{8n+4}-\frac{208}{8n+5}-\frac{256}{8n+6}+\frac{12}{8n+10}+\frac{26}{8n+12}+\frac{3}{8n+14}\right)=13(\pi-1)$$

$$\sum_{n=0}^{\infty}\frac{1}{16^{n+1}}\left(\frac{832}{8n+1}+\frac{192}{8n+2}-\frac{832}{8n+4}-\frac{208}{8n+5}-\frac{160}{8n+6}-\frac{12}{8n+10}+\frac{26}{8n+12}-\frac{3}{8n+14}\right)=13\pi$$

--

$$\sum_{n=0}^{\infty}\frac{1}{16^n}\left(\frac{4}{8n+1}-\frac{4}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}+\frac{1/8}{8n+12}\right)=\pi-\frac{1}{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{16^n}\left(\frac{4}{8n+1}-\frac{4}{8n+4}-\frac{1}{8n+5}-\frac{a}{8n+6}+\frac{b}{8n+12}\right)=\ln\left(\frac{3^{a+2-8b}}{5^{1-8b}}\right)-2a\cot^{-1}(2)+2\cot^{-1}(3)-4b+\frac{\pi}{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{16^n}\left(\frac{4}{8n+1}-\frac{4}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}+\frac{1/8-a}{8n+12}\right)=4a-8a\ln\frac{5}{3}+\pi-\frac{1}{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{16^n}\left(\frac{16}{8n+4}-\frac{1}{8n+12}\right)=4$$

---

$$\sum_{n=0}^{\infty}\frac{1}{16^n}\left(\frac{4}{8n+1}+\frac{4}{8n+2}-\frac{4}{8n+4}-\frac{1}{8n+5}\right)=\pi+\ln\left(\frac{3^3}{5}\right)$$

---

$$\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{2^{4n}}\left(\frac{4}{4n+1}+\frac{1}{4n+3}\right)=\ln{3}$$

$$\frac{1}{8}\sum_{n=0}^{\infty}\frac{1}{2^{4n}}\left(\frac{4}{4n+2}-\frac{1}{4n+4}\right)=\ln{\frac{5}{4}}$$

$$\frac{1}{16}\sum_{n=0}^{\infty}\frac{1}{2^{4n}}\left(\frac{4}{4n+3}+\frac{1}{4n+5}\right)=\ln{3}-1$$

$$\frac{1}{64}\sum_{n=0}^{\infty}\frac{1}{2^{4n}}\left(\frac{4}{4n+5}+\frac{1}{4n+7}\right)=\ln{3}-\frac{13}{12}$$

-

$$\sum_{n=1}^{\infty}\left[\frac{2^{2n}}\right]^{-3}\frac{6\left(n-\frac{1}{4}\right)-\frac{1}{2}}{n^3\cdot2^{2n}}=\pi^2$$

$$\sum_{n=1}^{\infty}\left[\frac{2^{2n}}\right]^{-3}\frac{6\left(n-\frac{1}{3}\right)}{n^3\cdot2^{2n}}=\pi^2$$

$$\sum_{n=1}^{\infty}\left[\frac{2^{2n}}\right]^{-3}\frac{6\left(n-\frac{7}{18}\right)+\frac{1}{3}}{n^3\cdot2^{2n}}=\pi^2$$

--

$$\sum_{n=1}^{\infty}\left[\frac{2^{2n}}\right]^{-3}\frac{6\left(n-\frac{1}{4}\right)-\frac{1}{2}}{n^3\cdot(-2)^{3n}}=-4G$$

$$\sum_{n=1}^{\infty}\left[\frac{2^{2n}}\right]^{-3}\frac{6\left(n-\frac{1}{3}\right)}{n^3\cdot(-2)^{3n}}=-4G$$

$$\sum_{n=1}^{\infty}\left[\frac{2^{2n}}\right]^{-3}\frac{6\left(n-\frac{7}{18}\right)+\frac{1}{3}}{n^3\cdot(-2)^{3n}}=-4G$$

--

$$k\ge0$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}{\frac{2k+1}{2}\choose n}=\frac{\pi}{2}\cdot\frac{(2k+1)!!}{(2k+2)!!}$$

-

$$\lim_{x \to \infty}\sum_{n=1}^{\infty}{2n \choose n}\frac{\zeta(2n)}{n^x}=2\zeta(2)$$

$$\lim_{x \to \infty}\sum_{n=1}^{\infty}{2n \choose n}\frac{\zeta(kn)}{n^x}=2\zeta(k)$$

$$\lim_{x \to \infty}\sum_{n=1}^{\infty}{2n \choose n}\frac{\zeta(kn)}{n^x}\cdot\frac{1}{j^{n-1}}=2\zeta(k)$$

$$\lim_{x \to \infty}\sum_{n=1}^{\infty}{2n \choose n}^t\frac{\zeta(kn)}{n^x}=2^t\zeta(k)$$

general: where $$(s,t,j)$$ are numbers

$$\lim_{x \to \infty}\sum_{n=1}^{\infty}{2n \choose n}^t\frac{\zeta(kn)}{n^x}\cdot\frac{1}{j^{n-1}}=2^t\zeta(k)$$

$$\lim_{x \to \infty}\sum_{n=1}^{\infty}{sn \choose n}^t\frac{\zeta(kn)}{n^x}\cdot\frac{1}{j^{n-1}}=s^t\zeta(k)$$

$$\lim_{n \to \infty}n\left(1-\frac{1}{e^{\alpha}}\left(1+\frac{1}{n}\right)^{\alpha n}\right)\prod_{k=1}^{m}\left(1\pm\frac{k^s}{n}\right)=\frac{\alpha}{2}\mp\sum_{k=1}^{m}k^s$$

$$\lim_{n \to \infty}n\left(1-\frac{1}{e^{\alpha}}\left(1+\frac{1}{n}\right)^{\alpha n}\right)\prod_{k=1}^{\infty}\left(1\pm\frac{k^{-s}}{n}\right)=\frac{\alpha}{2}\mp\zeta(s)$$

$$\lim_{n \to \infty}\frac{n}{e}\left(1+\frac{1}{n}\right)^{ n}\left[\prod_{k=1}^{\infty}\left(1+\frac{k^{-s}}{n}\right)-\prod_{k=1}^{\infty}\left(1-\frac{k^{-s}}{n}\right)\right]=2\zeta(s)$$

$$\lim_{n \to \infty}n^2\left(1-\frac{1}{e^{2\alpha}}\left(1+\frac{1}{n}\right)^{ 2\alpha n}\right)^2\prod_{k=1}^{\infty}\left(1-\frac{k^{-2s}}{n^2}\right)=\alpha^2-\zeta^2(s)$$

-

$$\lim_{n \to \infty}\sum_{k=0}^{n}\sum_{j=0}^{x}(-1)^j{x \choose j}\frac{k^j}{n^{j+1}}=\frac{1}{x+1}$$

$$\lim_{n \to \infty}2\sqrt{n}-\sum_{j=1}^{n}\frac{1}{\sqrt{j+1}}=1-\zeta\left(\frac{1}{2}\right)$$

$$\sum_{n=0}^{\infty}\frac{a^{4n}}{(4n)!}=\frac{\cos(a)+\cosh(a)}{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{(10n)!}=\frac{\cosh(1)}{5}+\frac{2}{5}\cosh\left(\frac{\phi}{2}\right)\cos\sqrt{\frac{\sqrt{5}}{4\phi}}+\frac{2}{5}\cosh\left(\frac{1}{2\phi}\right)\cos\sqrt{\frac{\phi\sqrt{5}}{4}}$$

$$-\frac{e^2}{2}+\frac{5}{2}\sum_{n=0}^{\infty}\frac{2^{5n}}{(5n)!}=e^{-\phi}\cos\sqrt{\phi^{-1}\sqrt{5}}+e^{\phi^{-1}}\cos\sqrt{\phi\sqrt{5}}$$ -

$$\sum_{k=0}^{N}\frac{(-1)^k{N \choose k}}{(k+1)(k+2)(k+3)...(k+j)}=\frac{1}{(j-1)!(N+j)}$$

-

$$\prod_{k=1}^{\infty}e^{-2}\left(1+\frac{1}{k}\right)^{k+1/2}\left(\frac{2k+1}{2k-1}\right)^k=\frac{e^2}{2}\cdot \frac{1}{\sqrt{\pi e}}$$

$$\prod_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{k+1/2}\left(\frac{2k-1}{2k+1}\right)^k=\frac{e}{\sqrt{\pi e}}$$

$$\lim_{n \to \infty}\sum_{j=0}^{an}\ln{an\choose j}=\frac{a^2}{2}$$

$$\lim_{x \to 1}(1-x)^{1/k}\sum_{n=0}^{\infty}x^{\left(\frac{n}{k}\right)^2}=\Gamma\left(\frac{1}{k}\right)$$

$$\lim_{x \to 1}(1-x)\left(\sum_{n=0}^{\infty}x^{\left(\frac{n}{2}\right)^2}\right)^2=\pi$$

--

$$\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(n+1\right)}=\sqrt[4]{e}\sqrt{\pi}\left[erf\left(1/2\right)-1\right]$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma\left(\frac{n+1}{2}\right)+2\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(n+1\right)}=-2$$

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(n+1\right)}\right)^2=\pi[L_o(1/2)-I_o(1/2)]$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma^2\left(\frac{n+1}{2}\right)}{\Gamma\left(n+1\right)}=\frac{4\pi}{\sqrt{3^3}}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\Gamma\left(n+\frac{a}{b}\right)}{\Gamma\left(n+1\right)}=\frac{\Gamma\left(\frac{a}{b}\right)}{2^{a/b}}$$

---

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\frac{1}{n(4n+1)^2}=-\sqrt{\pi}+\left(\frac{\pi}{4}\right)^2\cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$

---

$$\lim_{n \to \infty}\sum_{k=1}^{n}\Gamma^{-ka}\left(\frac{k}{n}\right)=\frac{e^{a\gamma}}{e^{a\gamma}-1}$$

$$\lim_{n \to \infty}\sum_{k=1}^{n}(-1)^k\Gamma^{-ka}\left(\frac{k}{n}\right)=\frac{e^{a\gamma}}{e^{a\gamma}+1}$$

-

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\Gamma^{2^ak+1-2^a}\left(\frac{2^ak+1-2^a}{2^an}\right)}=\frac{e^{\gamma}}{e^{2^a\gamma}-1}$$

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{(-1)^k}{\Gamma^{2^ak+1-2^a}\left(\frac{2^ak+1-2^a}{2^an}\right)}=\frac{e^{\gamma}}{e^{2^a\gamma}+1}$$

---

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\left[\Gamma\left(\frac{k-1+2^{-a}}{n}\right)\right]^{2^a(k-1)+1}}=\frac{e^{\gamma}}{e^{2^a\gamma}-1}$$

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\left[\Gamma\left(\frac{k-1+2^{-a}}{n}\right)\right]^{k-1+2^{-a}}}=\frac{e^{\gamma/2^a}}{e^{\gamma}-1}$$

--

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}\frac{1}{(n+1)^s}=2[\zeta(s+1)-1]$$

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}\frac{1}{n+2}\frac{1}{a^n}=2a-1-2a(a-1)\ln\left(\frac{a}{a-1}\right)$$

-

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(n+1\right)}\frac{1}{(4n+1)^2}=-\frac{\sqrt{\pi}}{2}+ \frac{\pi(4+\pi)}{64}\cdot\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(n\right)}\frac{1}{(4n-1)^2}=-\frac{\sqrt{\pi}}{2}+ \frac{\pi(8+3\pi)}{16}\cdot\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\frac{1}{(4n-1)^2}=\frac{\pi^2}{16}\cdot \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\frac{1}{(4n+1)^2}=\frac{\pi(4-\pi)}{64}\cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$

---

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\frac{1}{(4n-1)^3}=\frac{\pi}{128}\cdot \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}(16G-\pi^2)$$

$$\sum_{n=1}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\frac{1}{(4n+1)^3}=\frac{\pi}{512}\cdot \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}[\pi(8-\pi)-16G)]$$

---

$$\lim_{x \to 0}\sum_{n=1}^{\infty}\frac{\left(-\frac{1}{a}\right)^{n-1}}{(n^x+b)^c}=\frac{1}{(a+1)(b+1)^c}$$

$$\lim_{x \to 0}\sum_{n=1}^{\infty}\frac{\left(\frac{1}{a}\right)^{n-1}}{(n^x+b)^c}=\frac{1}{(a-1)(b+1)^c}$$

$$\lim_{x \to 0}\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\frac{\left(-\frac{1}{j}\right)^{n}}{n^x+j}=1-\zeta(2)$$

---

$$\sum_{n=2}^{\infty}\frac{\ln(n!)}{n!^2}=\ln(3)\ln(\pi)$$

---

$$\lim_{x \to \infty}\left(\frac{1}{n}\sum_{k=1}^{n}(k+1)^{z/x}\right)^{nx}=(n+1)!^z$$

$$\lim_{x \to 0}\left(\frac{1}{n}\sum_{k=1}^{n}(k+a)^{z/x}\right)^{x}=(n+a)^{z}$$

$$\lim_{x \to 0}\left(\frac{1}{n}\sum_{k=1}^{n}k^{1/x}\right)^{x}=n$$

--

$$\prod_{n=1}^{\infty}\left(x^n\right)^{j^{-n}}=\frac{x^j}{(j-1)^2}$$

$$\prod_{n=1}^{\infty}\left(\frac{\Gamma(2^n+\frac{1}{2})}{a^n\Gamma(2^n)}\right)^{2^{-n}}=\frac{8}{a^2}\frac{\sqrt{\pi}}{e^2}$$

---

$$\lim_{n \to \infty}\left(n^2-\frac{1}{\tan^2\left(\frac{1}{n}\right)}\right)=\frac{2}{3}$$

$$\prod_{n=0}^{\infty}\frac{(4n+2)^4(4n+4)^4}{(4n+1)(4n+3)^6(4n+5)}=2\left(\frac{\pi}{\Gamma\left(\frac{1}{4}\right)}\right)^4$$

$$\prod_{n=0}^{\infty}\frac{(6n+2)^4(6n+4)^4}{(6n+1)(6n+3)^6(6n+5)}=\frac{9}{8}$$

$$\prod_{n=0}^{\infty}\frac{(5n+2)^4(5n+4)^4}{(5n+1)(5n+3)^6(5n+5)}=\frac{(2\pi)^2}{\phi^5\sqrt{5}}\left(\frac{\Gamma\left(\frac{1}{5}\right)}{\Gamma^2\left(\frac{2}{5}\right)}\right)^5$$

$$\prod_{n=0}^{\infty}\frac{(3n+2)^5(3n+4)^{10}(3n+6)}{(3n+1)(3n+3)^{10}(3n+5)^5}=\frac{6^5}{\Gamma^9\left(\frac{1}{3}\right)}$$

$$\prod_{n=0}^{\infty}\frac{(5n+2)^5(5n+4)^{10}(5n+6)}{(5n+1)(5n+3)^{10}(5n+5)^5}=\frac{5}{\phi^{10}}\left(\frac{\Gamma^2\left(\frac{1}{5}\right)}{\Gamma^3\left(\frac{2}{5}\right)}\right)^5$$

$$\prod_{n=0}^{\infty}\frac{(6n+2)^7(6n+4)^{35}(6n+6)^{21}(6n+8)}{(6n+1)(6n+3)^{21}(6n+5)^{35}(6n+7)^{7}}=\frac{4\pi^{24}}{9\Gamma^{27}\left(\frac{1}{3}\right)}$$

---

$$\lim_{n \to \infty}n\sum_{j=1}^{n}(-1)^j\sin\left(\frac{n}{n^2+\alpha j^2}\right)=-\frac{\alpha}{2(1+\alpha)}$$

$$\lim_{n \to \infty}\left(\frac{an+b}{an-b}\right)^n=e^{\frac{2b}{a}}$$

-

$$\sum_{n=1}^{\infty}\left(\frac{4}{n+1}-\frac{2}{n+2}-\frac{1}{2n-1}-\frac{1}{2n+1}-2\ln\frac{2n+1}{2n}\right)=\gamma+\ln\frac{\pi}{16}$$

--

$$\sum_{n=1}^{\infty}\frac{\sinh^j(a)}{\cosh(2na)-\cosh(a)}=S_j$$

$$\sum_{n=1}^{\infty}\frac{\sinh(a)}{\cosh(2na)-\cosh(a)}=S_1=\frac{1}{e^a-1}$$

$$\sum_{n=1}^{\infty}\frac{\sinh^2(a)}{\cosh(2na)-\cosh(a)}=S_2=\frac{1+e^a}{2e^a}$$

$$\sum_{n=1}^{\infty}\frac{\sinh^3(a)}{\cosh(2na)-\cosh(a)}=S_3=\frac{e^a-1}{e^{2a}}\left(\frac{1+e^a}{2e^a}\right)^2=e^{-2a}\frac{S^2_2}{S_1}$$

$$\sum_{n=1}^{\infty}\frac{\sinh^4(a)}{\cosh(2na)-\cosh(a)}=S_4=\frac{(e^a-1)^2}{e^{3a}}\left(\frac{1+e^a}{2e^a}\right)^3=e^{-3a}\frac{S^3_2}{S^2_1}$$

$$\sum_{n=1}^{\infty}\frac{\sinh^5(a)}{\cosh(2na)-\cosh(a)}=S_5=\frac{(e^a-1)^3}{e^{4a}}\left(\frac{1+e^a}{2e^a}\right)^4=e^{-4a}\frac{S^4_2}{S^3_1}$$

$$S_{j+1}=e^{-aj}\frac{S^j_2}{S^{j-1}_1}$$

---

$$\sum_{n=1}^{\infty}\frac{\cosh^j(a)}{\cosh(2na)-\cosh(a)}=\frac{2e^a\cosh^{j}(a)}{(e^a-1)^2(e^a+1)}=\frac{S^2_1}{S_2}\cosh^{j}(a)$$

---

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n-1)n}{2}}\frac{1}{(2n+1)}=\frac{\pi}{2\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{n(n+1)}{2}}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n+1)(n+2)}{2}}\frac{1}{(2n+1)^3}=-\frac{3\pi^3}{64\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n+2)(n+3)}{2}}\frac{1}{(2n+1)^4}=-\frac{11\pi^4}{768\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n+3)(n+4)}{2}}\frac{1}{(2n+1)^5}=\frac{19\pi^5}{4096\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n+k-2)(n+k-1)}{2}}\frac{1}{(2n+1)^k}=\pi^kF(k)$$

and

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n-1)(n-2)}{2}}\frac{1}{(2n-1)}=1+\frac{\pi}{2\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{n(n-1)}{2}}\frac{1}{(2n-1)^2}=1+\frac{\pi^2}{8\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{n(n+1)}{2}}\frac{1}{(2n-1)^3}=-1-\frac{3\pi^3}{64\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n+1)(n+2)}{2}}\frac{1}{(2n-1)^4}=-1-\frac{11\pi^4}{768\sqrt{2}}$$

$$\sum_{n=0}^{\infty}(-1)^{\frac{(n+2)(n+3)}{2}}\frac{1}{(2n-1)^5}=1+\frac{19\pi^5}{4096\sqrt{2}}$$

--

$$\lim_{n\to \infty}n\left[e-\left(1+\frac{1}{n}\right)^n\right]\left[e-\left(1+\frac{a}{n}\right)^n\right]=\frac{e}{2}(e-e^a)$$

$$\lim_{n\to \infty}n\left[e-\left(1+\frac{1}{n}\right)^n\right]\left[e-\left(1+\frac{a}{n}\right)^n\right]\left[e-\left(1+\frac{b}{n}\right)^n\right]=\frac{e}{2}(e-e^a)(e-e^b)$$

and so on...

$$\lim_{n\to \infty}\prod_{j=1}^{n}\left(\frac{n^3+aj}{n^3-aj}\right)^n=e^a$$

$$\lim_{n\to \infty}e^{-an}\prod_{j=1}^{n}\left(\frac{n^2+aj}{n^2-aj}\right)^n=e^a$$

-

$$\prod_{n=0}^{\infty}\frac{1}{e}\left(\frac{2n+1}{2n-1}\right)^n=\frac{1}{\sqrt{2e}}$$

$$\prod_{n=2}^{\infty}\left(\frac{2n+1}{2n-1}\right)^n\left(\frac{n^2-1}{n^2}\right)^{n^2}=\frac{\pi}{3\sqrt{2}}$$

$$\sum_{n=1}^{\infty}\left(-\frac{1}{a}\right)^n\left(\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}\right)^{-1}=\frac{3a(a^2-6a+1)}{(1+a)^4}$$

$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-2\ln\frac{2n+1}{2n}\right)=\gamma+\ln\frac{\pi}{4}$$

$$\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}-2\ln\frac{2n+1}{2n}\right)=1+\gamma+\ln\frac{\pi}{16}$$

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1}\left(\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}\right)=G-\frac{\pi}{4}+\ln{2}-\frac{1}{2}$$

---

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^3}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)^2=8(G-\ln{2})$$

$$\sum_{n=0}^{\infty}\frac{1}{2n+1}\left[\frac{1}{(4n+1)^2}+\frac{1}{(4n+3)^2}\right]=2G-\ln{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{1}{4n+1}-\frac{1}{4n+3}\right)^2=2G-2\ln{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)^2=2G$$

$$\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)=\frac{\pi}{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{1}{4n+1}-\frac{1}{4n+3}\right)=\ln{2}$$

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)=\ln{4}$$

--

$$\sum_{n=0}^{\infty}\frac{2}{2n+1}\left(\frac{2}{4n+1}+\frac{2}{4n+3}\right)^3=\pi^3+2\pi^2-4\pi$$

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)^3=\frac{7}{2}\zeta(3)-2G$$

$$\sum_{n=0}^{\infty}\frac{2}{(2n+1)^3}\left(\frac{2}{4n+1}+\frac{2}{4n+3}\right)^3=4(\pi^3-6\pi^2+12\pi)$$

--

$$\int_{0}^{n}\pi(x)\mathrm dx=I(n)$$

$$I(2n+1)-2I(2n+2)+I(2n+3)=0$$

$$I(6n-5)-3I(6n-4)+3I(6n-3)-I(6n-2)=0$$

$$I(6n+1)-4I(6n+2)+6I(6n+3)-4I(6n+4)+I(6n+6)=0$$

- $$S(n)=\sum_{j=2}^{n-1}\pi(j)$$

$$S(2n+1)-2S(2n+2)+S(2n+3)=0$$

$$S(6n-5)-3S(6n-4)+3S(6n-3)-S(6n-2)=0$$

$$S(6n+1)-4S(6n+2)+6S(6n+3)-4S(6n+4)+S(6n+6)=0$$ --

$$\lim_{M \to \infty}\sum_{n=0}^{\infty}\sum_{k=0}^{2M-1}(-1)^k\zeta(2n+k+2)=\frac{\zeta(2)+\frac{1}{2}}{2}$$

--

$$\sum_{n=0}^{\infty}\left[\zeta(2n)-2\zeta(2n+3)+\zeta(2n+5)\right]=-\zeta(3)$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n+4)-\zeta(2n+5)\right]=\zeta(3)-\zeta(2)+\frac{1}{2}$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n+4)-\zeta(2n+3)\right]=-\zeta(2)+\frac{3}{2}$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n+2)-\zeta(2n+3)\right]=\frac{1}{2}$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n)-\zeta(2n+2)\right]=-\frac{3}{2}$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n)-\zeta(2n+3)\right]=-1$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n+2)-\zeta(2n+4)\right]=\zeta(2)-1$$

$$\sum_{n=0}^{\infty}\left[\zeta(2n+2)-\zeta(2n+5)\right]=\zeta(3)-\frac{1}{2}$$

--

$$\int_{0}^{n}\pi(x)\pi(n-x)\mathrm dx=I(n)$$

$$I(10n+6)-2I(10+7)+I(10+8)=0$$

$$k\ge1$$,

$$I[2(2k+1)n+2k+2]-2I[2(2k+1)n+2k+3]+I[2(2k+1)n+2k+4]=0$$

--

$$\lim_{j \to \infty}\prod_{k=1}^{\infty}\left[\prod_{n=j}^{2j}\left(1-{\ln\left(1+\frac{1}{k}\right)^{\frac{(-1)^{k}}{n}}}\right)\right]^{\frac{1}{\ln{2}}}=\frac{\pi}{2}$$

$$\lim_{j,M \to \infty}\frac{1}{M}\prod_{k=1}^{M}\left[\prod_{n=j}^{2j}\left(1+\frac{1}{kn}\right)\right]^{\frac{1}{\ln{2}}}=e^{\gamma}$$

$$\lim_{j \to \infty}\prod_{k=1}^{\infty}\left[\prod_{n=j}^{2j}\left(1-{\ln\left(1+\frac{1}{k}\right)^{\frac{(-1)^{k}}{n}}}\right)\right]^{\frac{2j}{n}}=\frac{\pi}{2}$$

$$\lim_{j,M \to \infty}\frac{1}{M}\prod_{k=1}^{M}\left[\prod_{n=j}^{2j}\left(1+\frac{1}{kn}\right)\right]^{\frac{2j}{n}}=e^{\gamma}$$

$$\lim_{j \to \infty}\prod_{k=1}^{\infty}\left[\prod_{n=j}^{2j}\left(1+\frac{f(k)}{n}\right)\right]^{\frac{2j}{n}}=e^{\sum_{k=1}^{\infty}f(k)}$$

$$\lim_{j \to \infty}\prod_{k=1}^{\infty}\left[\prod_{n=j}^{kj}\left(1+\frac{f(k)}{n}\right)\right]^{\frac{1}{\ln{k}}}=e^{\sum_{k=1}^{\infty}f(k)}$$

$$\lim_{j \to \infty}\sum_{n=j}^{kj}(-1)^n\frac{(jk)^2}{n(k^2+1)}\ln\left(1+\frac{\alpha}{n}\right)=\frac{\alpha}{2}$$

-

$$\lim_{(j,m) \to \infty}\sum_{n=j}^{kj}\frac{jk}{n(k-1)}\prod_{x=1}^{m}\ln\left(1+\frac{1}{xn}\right)-\ln(m)=\gamma$$

-

$$\lim_{j \to \infty}\sum_{n=j}^{kj}\frac{1}{n}=\ln{(k)}$$

$$\lim_{j \to \infty}\sum_{n=j}^{kj}\ln\left(1+\frac{\alpha}{n}\right)=\alpha\ln{(k)}$$

$$\lim_{j \to \infty}\sum_{n=j}^{kj}\ln\left(\frac{n+\alpha}{n-\beta}\right)=(\alpha+\beta)\ln{(k)}$$

$$\lim_{j \to \infty}\sum_{n=j}^{kj}\frac{jk}{n(k-1)}\ln\left(1+\frac{\alpha}{n}\right)=\alpha$$

$$\lim_{j \to \infty}\sum_{n=j}^{kj}\frac{jk}{n(k-1)}\ln\left(\frac{n+\alpha}{n-\beta}\right)=\alpha+\beta$$

$$\lim_{(j,m) \to \infty}\sum_{n=j}^{kj}\frac{jk}{n(k-1)}\prod_{x=1}^{m}\ln\left(1+\frac{f(x)}{n}\right)=\sum_{n=1}^{\infty}f(x)$$

$$\lim_{(j,m) \to \infty}\sum_{n=j}^{kj}\frac{jk}{n(k-1)}\prod_{x=1}^{m}\ln\left(\frac{n+f(x)}{n-g(x)}\right)=\sum_{n=1}^{\infty}[f(x)+g(x)]$$

---

$$\sum_{n=1}^{\infty}\left[\frac{3}{2n}+\frac{1}{2n-1}+2\ln\left(\frac{2n-1}{2n+1}\right)\right]=2\gamma-\ln{2}$$

$$\sum_{n=1}^{\infty}\left[\frac{1}{2n}-\frac{1}{2n-1}+2\ln\left(\frac{2n}{2n+1}\cdot\frac{2n}{2n-1}\right)\right]=\ln\frac{\pi^2}{8}$$

$$\sum_{n=1}^{\infty}\left[\frac{2}{2n-1}+\ln\frac{(2n+1)^2(2n-1)^4}{(4n^2)^3}\right]=\gamma-\ln\frac{\pi^3}{2^4}$$

$$\sum_{n=1}^{\infty}\left[\frac{3}{2n}-\frac{1}{2n-1}-\ln\frac{(2n-1)^2(2n+1)^4}{(4n^2)^3}\right]=\gamma+\ln\frac{\pi^3}{2^5}$$

$$k\ge1$$

$$\prod_{n=2k}^{\infty}\left[\frac{n}{\left(1+\frac{1}{n}\right)^n}\right]^{\frac{1}{n(n+1)}}=2^{\gamma-\ln\left(1+\frac{1}{\sqrt{\phi^3}+\phi}\right)\left(1+\frac{1}{\sqrt{2\phi^5}+\phi^3}\right)}$$

$$\prod_{n=2k}^{\infty}\left[\frac{n}{\left(1+\frac{1}{n}\right)^n}\right]^{\frac{1}{n(n+1)}}=2^{\gamma-\ln{\sqrt{2}}}$$

$$\prod_{n=2k}^{\infty}\left[\frac{n^2}{\left(1+\frac{2}{n}\right)^n}\right]^{\frac{1}{n(n+2)}}=2$$

$$\prod_{n=2k}^{\infty}\left[\frac{n^{2n+1}}{\left(1+\frac{1}{n}\right)^{n^2}}\right]^{\frac{2}{(n^2+n)^2}}=2^{\zeta(2)}\cdot e^{\zeta^{'}(2)}$$

---

$$\prod_{n=1}^{\infty}\left[\frac{n}{\left(1+\frac{1}{n}\right)^n}\right]^{\frac{(-1)^n}{n(n+1)}}=2^{2\gamma-\ln{2}}$$

$$\prod_{n=1}^{\infty}\left[\frac{n^2}{\left(1+\frac{2}{n}\right)^n}\right]^{\frac{(-1)^n}{n(n+2)}}=\sqrt{2}$$

$$\prod_{n=1}^{\infty}\left[\frac{n^{2n+1}}{\left(1+\frac{1}{n}\right)^{n^2}}\right]^{\frac{(-1)^n}{(n^2+n)^2}}=2^{\zeta(2)}\cdot e^{\zeta^{'}(2)}$$

$$\prod_{n=1}^{\infty}\left[\frac{n^{4(n+1)}}{\left(1+\frac{2}{n}\right)^{n^2}}\right]^{\frac{(-1)^n}{(n^2+2n)^2}}=\sqrt[4]{2}$$

-

$$\sum_{n=1}^{\infty}(-1)^{n+1}\cdot \frac{\ln(n+\alpha)}{(n+\alpha)^b}=f(\alpha,b)$$

$$F(0,1)=\frac{1}{2}\ln(2)[\ln(2)-2\gamma]$$

$$F(1,1)=-\frac{1}{2}\ln(2)[\ln(2)-2\gamma]$$

$$F(2,1)=\frac{1}{2}\ln(2)[\ln(2)-2\gamma+1]$$

note that

$$F(2\alpha,b)-f(\alpha,b)=$$ simple closed form

---

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n(1+n)}\cdot\ln\left(\frac{n}{(1+\frac{1}{n})^n}\right)=\ln2[2\gamma-\ln2]$$

-

$$\sum_{n=1}^{\infty}\left(H_{n}-H_{2n}-H_{4n}+H_{8n}-\frac{3}{16n}\right)=\frac{3-\pi\sqrt{2}}{16}$$

$$\sum_{n=1}^{\infty}\left[H_{n}-2H_{2n}+4H_{4n}-8H_{8n}+\ln(2^{18})+5(\gamma+\ln n)\right]=\frac{\pi}{4}(1+\sqrt{8})+\frac{5(\ln{\pi}-\gamma)-13\ln{2}}{2}$$

$$\sum_{n=1}^{\infty}\left(H_{2n}-\ln{(2n)}-\frac{1}{4n}-\gamma\right)=\frac{1}{4}+\frac{\gamma-\ln{\pi}}{2}$$

$$\sum_{n=1}^{\infty}\left(H_{4n}-\ln{(4n)}-\frac{1}{8n}-\gamma\right)=\frac{2-\pi}{16}+\frac{\gamma-\ln\frac{\pi}{2}}{2}$$

$$\sum_{n=1}^{\infty}\left(H_{8n}-\ln{(8n)}-\frac{1}{16n}-\gamma\right)=\frac{1-\pi(1+\sqrt{2})}{16}+\frac{\gamma-\ln\frac{\pi}{4}}{2}$$

---

$$\sum_{n=1}^{\infty}\left(H^{'}_{2n}-\ln{2}+\frac{1}{4n}\right)=\frac{\ln 2}{2}-\frac{1}{4}$$

$$\sum_{n=1}^{\infty}\left(H^{'}_{4n}-\ln{2}+\frac{1}{8n}\right)=\frac{\ln{2}}{2}-\frac{\pi+2}{16}$$

$$\sum_{n=1}^{\infty}\left(H^{'}_{8n}-\ln{2}+\frac{1}{16n}\right)=\frac{\ln{2}}{2}-\frac{1+\pi\sqrt{2}}{16}$$

-

$$\sum_{n=1}^{\infty}\left[H_{2n}+H^{'}_{2n}-\gamma-\ln(4n)\right]=\frac{\gamma-\ln\frac{\pi}{2}}{2} $$

$$\sum_{n=1}^{\infty}\left[H_{4n}+H^{'}_{4n}-\gamma-\ln(8n)\right]= \frac{4\left(\gamma-\ln\frac{\pi}{4}\right)-\pi}{8}$$

$$\sum_{n=1}^{\infty}\left[H_{8n}+H^{'}_{8n}-\gamma-\ln(16n)\right]= \frac{8\left(\gamma-\ln\frac{\pi}{8}\right)-\pi(1+\sqrt{8})}{16} $$

$$\sum_{n=1}^{\infty}\left[H_{n}-2H_{2n}-H_{4n}+2H_{8n}-\ln(4)\right]= \ln2-\frac{\pi}{16}(1+\sqrt{8}) $$

$$\sum_{n=1}^{\infty}\left[H_{n}-3H_{2n}+2H_{4n}-\ln(2)\right]= \frac{\ln2-\frac{\pi}{4}}{2} $$

-- $$\sum_{n=1}^{\infty}\left(H_{n}+H_{2n}+H_{4n}-3[\gamma+\ln{(2n)}]-\frac{7}{8n}\right)=\frac{14-\pi+24(\gamma-\ln{\pi})}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{n}+H_{2n}-H_{4n}-\ln\left(\frac{n}{2}\right)-\gamma-\frac{5}{8n}\right)=\frac{10+\pi+8(\gamma-\ln{(4\pi)})}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{n}-H_{2n}+H_{4n}-\ln{(2n)}-\gamma-\frac{3}{8n}\right)=\frac{6-\pi+8(\gamma-\ln{\pi})}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{n}-H_{2n}-H_{4n}+\ln{(8n)}+\gamma-\frac{1}{8n}\right)=\frac{2+\pi-8(\gamma-\ln\frac{\pi}{4})}{16}$$

$$\sum_{n=1}^{\infty}\left( H_{n}-2H_{2n}+\ln{4n}+\gamma\right)=\frac{\ln\frac{\pi}{2}-\gamma}{2}$$

$$\sum_{n=1}^{\infty}\left( H_{n}-H_{4n}+\ln{4}-\frac{3}{8n}\right)=\frac{6+\pi-16\ln{2}}{16}$$

$$\sum_{n=1}^{\infty}\left( H_{n}-4H_{4n}+3\ln{n}+3\gamma+8\ln{2}\right)=\frac{\pi+6\left(\ln\frac{\pi}{2^{5/3}}-\gamma\right)}{4}$$

-

$$\sum_{n=1}^{\infty}\left( H_{2n}-H_{4n}-\frac{1}{8n}+\ln{2}\right)=\frac{2(1-\ln{16})+\pi}{16}$$

$$\sum_{n=1}^{\infty}\left(2H_{2n}-H_{4n}-\ln{n}-\gamma-\frac{3}{8n}\right)=\frac{8(\gamma-\ln{2\pi})+\pi+6}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{2n}-2H_{4n}+\ln{n}+\gamma+3\ln{2}\right)=\frac{4\left(\ln{\frac{\pi}{4}}-\gamma\right)+\pi}{8}$$

$$\sum_{n=1}^{\infty}\left(H_{2n}+H_{4n}-2\ln{n}-2\gamma-3\ln{2}-\frac{3}{8n}\right)=\frac{16\left(\ln{\frac{\sqrt{2}}{\pi}}+\gamma\right)+6-\pi}{16}$$

--

$$\sum_{n=1}^{\infty}\left( H_{n}+H_{2n}-2H_{4n}-\frac{1}{2n}+\ln{8}\right)=\frac{4(1-\ln{8})+\pi}{8}$$

$$\sum_{n=1}^{\infty}\left(3H_{n}-8H_{2n}+4H_{4n}+\ln{n}+\gamma\right)=\frac{2[\ln{(2\pi)-\gamma}]-\pi}{4}$$

$$\sum_{n=1}^{\infty}\left(H_{n}-2H_{2n}+H_{4n}-\frac{1}{8n}\right)=\frac{2-\pi}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{n}+2H_{2n}-8H_{4n}+5\ln{n}+5\gamma+14\ln 2\right)=\frac{\pi-5\gamma+\ln\left(\frac{\pi^5}{2^9}\right)}{2}$$

$$\sum_{n=1}^{\infty}\left(H_{n}-4H_{2n}+4H_{4n}-\ln{n}-\gamma-4\ln 2\right)=\frac{2\gamma-\pi+2\ln\left(\frac{8}{\pi}\right)}{4}$$

--

$$\sum_{n=1}^{\infty}\left( H_{n}+H_{2n}-2H_{4n}-\frac{1}{2n}+\ln{8}\right)=\frac{4(1-\ln{8})+\pi}{8}$$

$$\sum_{n=1}^{\infty}(-1)^n\left(H_{n}-\ln{n}-\frac{1}{2n}-\gamma\right)=\frac{\gamma-\ln\frac{\pi}{2}}{2}$$

$$\sum_{n=1}^{\infty}\left(H_{2n}-\ln{(2n)}-\frac{1}{4n}-\gamma\right)=\frac{1+2\left(\gamma-\ln{\pi}\right)}{4}$$

$$\sum_{n=1}^{\infty}\left(H_{4n}-\ln{(4n)}-\frac{1}{8n}-\gamma\right)=\frac{2-\pi+8\left(\gamma-\ln\frac{\pi}{2}\right)}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{4n}-\ln{(4n-2)}-\frac{1}{4n-2}-\frac{3}{8n}-\gamma\right)=\frac{2-\pi}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{\alpha n}-\ln{(\alpha n)}-\frac{1}{2\alpha n}-\gamma\right)=F(\alpha)$$

$$\sum_{n=1}^{\infty}\left(H_{2n-1}-\ln{(2n-1)}-\frac{1}{2(2n-1)}-\gamma\right)=\frac{1-\ln{4}}{4}$$

$$\lim_{k \to \infty}\frac{1}{\sqrt{k}}\cdot\sum_{n=1}^{k}\frac{1}{\sum_{j=0}^{M}\sqrt{2n+j}}=\frac{1}{\sqrt{2}}\cdot\frac{1}{2^{M-1}}$$

$$\sum_{n=1}^{\infty}\left(\frac{1}{2n}+\frac{1}{2n-1}+2\ln\frac{2n-1}{2n}\right)=\gamma-\ln\frac{\pi}{2}$$

$$\sum_{n=1}^{\infty}\left(H_{n}-2H_{2n-1}+\ln{(4n-2)}+\frac{1}{4n-2}-\frac{3}{4n}+\gamma\right)=0$$

$$\sum_{n=1}^{\infty}\left(H_{4n}-H_{2n}+\frac{1}{8n}-\ln{2}\right)=\frac{-2-\pi+8\ln{2}}{16}$$

$$\sum_{n=1}^{\infty}\left(H_{4n}-H_{2n}-\ln\left(\frac{2n-1}{n}\right)-\frac{1}{4n-2}-\frac{1}{8n}\right)=-\frac{2+\pi+8(\gamma-\ln \pi)}{16}$$

$$\sum_{n=1}^{\infty}\left(2H_{4n}-2H_{2n}-\ln\left(\frac{4n-2}{n}\right)-\frac{1}{4n-2}\right)=-\frac{2+\pi+4(\gamma-\ln (2\pi))}{8}$$

--

$$2\sum_{n=0}^{\infty}(-1)^n\frac{2^n}{2^{2^n}-1}-\sum_{n=0}^{\infty}(-1)^n\frac{2^n}{2^{2^n}+1}=1$$

--

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{k}{a(j+k)}=\frac{1}{4a-1}+\frac{4a}{\sqrt{(4a-1)^3}}\cdot \sin^{-1}\left(\frac{1}{2\sqrt{a}}\right)$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{k}{a(j+k+1)}=\frac{1-2a}{4a-1}+\frac{8a^2}{\sqrt{(4a-1)^3}}\cdot \sin^{-1}\left(\frac{1}{2\sqrt{a}}\right)$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{k}{a(j+k-1)}=\frac{2}{4a-1}+\frac{8a}{\sqrt{(4a-1)^3}}\cdot \sin^{-1}\left(\frac{1}{2\sqrt{a}}\right)$$

---

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k-1}=2+\pi$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k}=\frac{\pi+2}{2}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=\frac{\pi}{2}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+2}=\frac{\pi-1}{2}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+3}=\frac{\pi}{4}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+4}=\frac{8-\pi}{8}$$

---

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2(k+1)}{j+k-1}=3\pi+8$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2(k+1)}{j+k}=\frac{3\pi+8}{2}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2(k+2)}{j+k+2}=\frac{3\pi+7}{2}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\left[1-\frac{2(k+1)}{j+k+1}\right]=-\frac{1}{3}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\left[1-\frac{2(k+1)}{j+k-1}\right]=-\frac{21}{10}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\left[1-\frac{1}{2}\cdot\frac{4k-1}{j+k+1}\right]=\frac{7}{9}$$

---

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\left[\frac{\left(k+\frac{1}{2}\right)}{j+k+1}\right]^2=\frac{977}{3375}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\left[\frac{\left(k+\frac{1}{2}\right)}{j+k-1}\right]^2=2\frac{2342}{3375}$$

-

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{\alpha\left(k+\frac{1}{2}\right)}{j+k+1}=F(\alpha)$$

$$F(1)=\frac{7}{9}$$, $$F(2)=3$$, $$F(3)=15$$, $$F\left(\frac{1}{4}\right)=\frac{31}{225}$$

---

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{\alpha\left(k+\frac{1}{2}\right)}{j+k-1}=F(\alpha)$$

$$F(1)=\frac{22}{9}$$, $$F(2)=10$$, $$F(3)=54$$, $$F\left(\frac{1}{4}\right)=\frac{94}{225}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2(k+1)}{j+k+1}=2+\pi$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2(k+1)}{j+k-1}=3\pi+8$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=\frac{\pi}{2}$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k+1}{j+k+1}=3$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k+1}{j+k-1}=10$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k-1}{j+k-1}=2$$

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{3k-2}{j+k+1}=1$$

$$\sum_{j=1}^{N}\prod_{k=1}^{j}\frac{2(2k-1)}{j+k+1}=2N$$

---

$$\sum_{n=1}^{\infty}(a)^n\cdot\frac{1}{LCM[{2n \choose n}n,n+1]+LCM[{2n \choose n}n,n-1]}=\sin^{-1}\left(\frac{\sqrt{a}}{2}\right)^2$$

$$\sum_{n=1}^{\infty}(-a)^n\cdot\frac{1}{LCM[{2n \choose n}n,n+1]+LCM[{2n \choose n}n,n-1]}=-\sinh^{-1}\left(\frac{\sqrt{a}}{2}\right)^2$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{5^n}\cdot\frac{1}{LCM[{2n \choose n},n+1]-LCM[{2n \choose n},n-1]}=\frac{1}{\phi}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{5^n}\cdot\frac{1}{LCM[{2n \choose n},n+1]+LCM[{2n \choose n},n-1]}=\ln\left(1+\frac{1}{\phi^2}\right)$$

$$\sum_{n=1}^{\infty}\frac{a^n}\cdot\frac{1}{LCM[{2n \choose n},n+1]-LCM[{2n \choose n},n-1]}=\frac{1}{2(a-1)}$$

$$\sum_{n=1}^{\infty}\frac{a^n}\cdot\frac{1}{LCM[{2n \choose n},n+1]+LCM[{2n \choose n},n-1]}=\frac{1}{2}\ln\left(\frac{a}{a-1}\right)$$

$$\sum_{n=1}^{\infty}\frac{1}{a^n}\cdot\frac{1}{LCM[{2n \choose n},n+1]+LCM[{2n \choose n},n-1]}=\frac{1}{\sqrt{4a-1}}\sin^{-1}\left(\frac{1}{2\sqrt{a}}\right)$$

$$\sum_{n=1}^{\infty}\frac{1}{a^n}\cdot\frac{1}{LCM[{2n \choose n},(n+1)^2]-LCM[{2n \choose n},(n-1)^2]}=\frac{1}{2\sqrt{4a-1}}\sin^{-1}\left(\frac{1}{2\sqrt{a}}\right)$$

--

$$\sum_{n=1}^{\infty}\frac{H_{n}}{2^n}\cdot\frac{1}{LCM[n+a,n]-LCM[n-b,n]}=\frac{\pi^2}{24(a+b)}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{2^n}\cdot\frac{1}{LCM[n+1,n^2]-LCM[n-1,n^2]}=\frac{\zeta(3)}{2}-\frac{\pi^2}{24}\cdot\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{n,2}}{2^n}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\frac{5\zeta(3)}{16}$$

$$\sum_{n=1}^{\infty}\frac{H_{n-1}}{2^n}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\frac{\ln^2{(2)}}{4}$$

$$\sum_{n=1}^{\infty}\frac{1}{a^n{2n \choose n}}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\frac{1}{\sqrt{a^2+a-1}}\sin^{-1}\left(\frac{1}{2\sqrt{a}}\right)$$

$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\frac{\pi}{4}$$

$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{LCM[n+1,n^2]-LCM[n-1,n^2]}=\frac{\pi^2}{16}$$

$$\sum_{n=1}^{\infty}\frac{a^n}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\sqrt{\frac{a}{4-a}}\sin^{-1}\left(\frac{\sqrt{a}}{2}\right)$$

$$\sum_{n=1}^{\infty}\frac{a^n}\cdot\frac{1}{LCM[n+1,n^2]-LCM[n-1,n^2]}=\sin^{-1}\left(\frac{\sqrt{a}}{2}\right)^2$$

$$\sum_{n=1}^{\infty}\frac{a^n}\cdot\frac{1}{\left(LCM[n+1,n^2]-LCM[n-1,n^2]\right)^2}=\frac{1}{2}\sin^{-1}\left(\frac{\sqrt{a}}{2}\right)^2$$

--

$$\sum_{n=1}^{\infty}\frac{4^n}\cdot\frac{1}{LCM[n+1,n]+LCM[n-1,n]}=\left(\frac{\pi}{2}\right)^2$$

$$\sum_{n=1}^{\infty}\frac{3^n}\cdot\frac{1}{LCM[n+1,n]+LCM[n-1,n]}=\left(\frac{\pi}{3}\right)^2$$

$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{LCM[n+1,n]+LCM[n-1,n]}=\left(\frac{\pi}{4}\right)^2$$

$$\sum_{n=1}^{\infty}\frac{1}\cdot\frac{1}{LCM[n+1,n]+LCM[n-1,n]}=\left(\frac{\pi}{6}\right)^2$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM[n+a,n^s]-LCM[n-b,n^s]}=\frac{\zeta(s)}{a+b}$$

$$\sum_{n=1}^{\infty}\frac{4^n}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM[n+1,n^s]-LCM[n,n^s]-LCM[n-1,n^s]}=\zeta(s)$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM[n+3,n^s]-LCM[n+2,n^s]-LCM[n+1,n^s]}=-\zeta(s+1)$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM[n+4,n^s]-LCM[n+2,n^s]-LCM[n,n^s]}=\zeta(s)$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM[n+6,n^s]-LCM[n+4,n^s]-LCM[n+2,n^s]}=-\zeta(s+1)$$

$$\sum_{n=1}^{\infty}\frac{{2n\choose n}^2}{16^n}\cdot\frac{1}{LCM[n+1,n]-LCM[n-1,n]}=\ln{4}-\frac{4G}{\pi}$$

$$\sum_{n=1}^{\infty}\frac{{2n\choose n}^2}{16^n}\cdot\frac{1}{LCM[n+a,n]-LCM[n-b,n]}=\frac{2}{a+b}\left(\ln{4}-\frac{4G}{\pi}\right)$$

--

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{LCM\left({2n \choose n},n^2\right)}=-2\ln^2{(\phi)}$$

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{LCM\left({2n \choose n},n\right)}=-\frac{2\ln{(\phi)}}{\sqrt{5}}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{LCM\left({2n \choose n},2n+1\right)}=\frac{4\ln{(\phi)}}{\sqrt{5}}$$

-

$$\sum_{n=1}^{\infty}\frac{1}{LCM^3(n,n+1)}=10-\pi^2$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM\left({2n \choose n},n^2\right)}=\frac{\pi^2}{18}$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM\left({2n \choose n},n\right)}=\frac{\pi}{\sqrt{27}}$$

$$\sum_{n=0}^{\infty}\frac{1}{LCM\left({2n \choose n},2n+1\right)}=\frac{2\pi}{\sqrt{27}}$$

--

$$\sum_{n=1}^{\infty}\frac{1}{LCM(kn,2n+1)}=\frac{2}{k}(1-\ln{2})$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM(n+k,n+k+1,n+k+2)}=\frac{1}{2(k+1)(k+2)}$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM(n^2,(n+2)(n+3))}=-\frac{1}{6}+\frac{\pi^2}{6^2}-\frac{1}{6^3}$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM(n+1,n^s)}=(-1)^s[-1+\zeta(2)-\zeta(3)+\cdots +\zeta(s)]$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM(n,(n+1)^s)}=s-\zeta(2)-\zeta(3)-\cdots -\zeta(s)$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM(n+1,(n+1)^s)}=\zeta(s)-1$$

$$\sum_{n=1}^{\infty}\frac{1}{LCM(n+2,(n+1)^s)}=\frac{1}{2}+(-1)^s\left[-1+\zeta(2)-\zeta(3)+\cdots +\zeta(s)\right]$$

-

$$\sum_{n=1}^{\infty}\frac{1}{n(2n-1)(4n-1)}=\pi-4\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{8n^2+1}{n(4n^2-1)(16n^2-1)}=3-4\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{8n^2+7}{n(4n^2-1)(16n^2-1)}=17-24\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{8n^2+6k+1}{n(4n^2-1)(16n^2-1)}=14k+3-4(5k+1)\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{40n^2-1}{n(4n^2-1)(16n^2-1)}=1$$

$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)(4n+1)}=6-\pi-4\ln{2}$$

$$\sum_{n=1}^{\infty}\frac{1}{n(4n-1)}=3\ln{2}-\frac{\pi}{2}$$

$$\sum_{n=1}^{\infty}\frac{1}{n(4n-3)}=\ln{2}+\frac{\pi}{6}$$

$$\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)(16n^2-1)}=\frac{\pi-3}{6}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{(4n^2-1)(16n^2-1)}=\frac{1-\ln{4}}{10}$$

$$\sum_{n=0}^{\infty}\frac{n}{(4n^2-1)(16n^2-1)}=\frac{1+\ln{16}-\ln{27}}{10}$$

---

$$\sum_{n=1}^{\infty}\frac{8n-3}{n(4n-1)(4n-3)}=\ln{8}$$

$$\sum_{n=-\infty}^{\infty} (-1)^n e^{-\pi n^2}=\left(\frac{\pi}{2}\right)^{1/4}\cdot\frac{1}{\Gamma\left(\frac{3}{4}\right)}$$

--

$$\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^{2k+1}=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^{2k+1}$$, $$k\ge0$$

--

$$\sum_{n=1}^{\infty}\frac{\sin^4(22n)}{7n^2}=\frac{\pi}{2}\left(\frac{22}{7}-\pi\right)$$

$$\sum_{n=1}^{\infty}\frac{\sin^4(22n)\cos^4(22n)}{7n^2}=\frac{\pi}{2^5}\left(\frac{22}{7}-\pi\right)$$

$$\sum_{n=1}^{\infty}\frac{\sin^{2(k+1)}(22n)}{n^{2k}}=\frac{\pi}{4k}\left(22-7\pi\right)^{2k-1}$$, $$k\ge 1$$

-

$$k\ge0$$

$$\sum_{n=1}^{\infty}\frac{\sin(22n)^{2k+4}}{7n^2}=\frac{(2k+1)!!}{(2k+2)!!}\cdot \frac{\pi}{2}\left(\frac{22}{7}-\pi\right)$$

$$\sum_{n=1}^{\infty}\frac{\sin[2n(3k-1)]^{2k+4}}{n^2}=\frac{(2k+1)!!}{(2k+2)!!}\cdot \frac{\pi}{2}\left[2(3k-1)-\pi(2k-1)\right]$$

-

$$\sum_{n=1}^{\infty}\left(\frac{\sin(kn)}{n}\right)^2=-\frac{1}{2}(\alpha \pi-k)(\beta \pi-k)$$

$$\sum_{n=1}^{\infty}\left(\frac{\sin(kn)}{n}\right)^3=(\alpha \pi-\beta)^2(a\pi-b)$$

$$\sum_{n=1}^{\infty}\left(\frac{\sin(4n)}{n}\right)^3=\frac{1}{2}(\pi-4)^3$$

$$\sum_{n=1}^{\infty}\left(\frac{\sin[2n(3k-1)]}{n}\right)^3=\frac{1}{2}[(2k-1)\pi-2(3k-1)]^3$$

$$\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^3$$

$$\frac{\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3}{\left(\pi-\frac{22}{7}\right)^3}=\frac{1}{2}$$

--

$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{X^{2(k+1)n}}=G(k,X)$$

$$2\sum_{n=1}^{\infty}\frac{\zeta(2n)}{X^{2n-1}}=X-\pi\cot\left(\frac{\pi}{X}\right)$$

---

$$\frac{\sqrt{2}}{2^{-2}}\cdot\frac{\sqrt{2-\sqrt{2}}}{2^{-3}}\cdot \frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2^{-4}}\cdot\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2^{-5}}\cdots\prod_{k=1}^{\infty}\frac{1}{\pi}\left(\frac{1}{2}-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{(k+1)n}}\right)=\frac{2}{\pi} $$

$$\frac{1}{2}-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{4n}}=\frac{\pi}{2^3}\cdot\frac{\sqrt{2}}{\sqrt{2}}$$

$$\frac{1}{2}-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{6n}}=\frac{\pi}{2^4}\cdot\frac{\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}$$

$$\frac{1}{2}-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{8n}}=\frac{\pi}{2^5}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2-\sqrt{2+\sqrt{2}}}}$$

$$\frac{1}{2}-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{10n}}=\frac{\pi}{2^6}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}$$

--

$$\frac{2}{\pi}=\frac{1}{2^k}\cdot\frac{\sqrt{2_1+\sqrt{2_2+\sqrt{2_3+\sqrt{2+\cdots+\sqrt{2_k}}}}}}{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cdots \sqrt{2}}}}}}$$

---

$$\sum_{n=1}^{\infty}n\cdot\frac{\zeta(2n)}{2^{2n-1}}=\frac{\pi^2}{8}$$

$$\sum_{n=1}^{\infty}n^2\cdot\frac{\zeta(2n)}{2^{2n-1}}=\frac{3\pi^2}{16}$$

$$\sum_{n=1}^{\infty}\frac{(2n-1)^2}{2^{2n-1}}\zeta(2n)=1+\frac{\pi^2}{4}$$

-

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{{(-1)^n}L_{n+1}+L_{n+2}}\right)^{(-1)^n}=\frac{2}{5}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{{(-1)^n}L_{n+1}+L_{n+4}}\right)^{(-1)^n}=\frac{8}{9}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{{(-1)^n}L_{n+2}+L_{n+5}}\right)^{(-1)^n}=\frac{14}{15}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{{(-1)^n}L_{n+3}+L_{n+4}}\right)^{(-1)^n}=\frac{11}{15}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{{(-1)^n}L_{n+4}+L_{n+7}}\right)^{(-1)^n}=\frac{36}{37}$$

$$\prod_{n=1}^{\infty}\left(1+~\frac{1}{{(-1)^n}L_{n+5}+L_{n+6}}\right)^{(-1)^n}=\frac{319}{360}$$ -

$$\sum_{n=1}^{\infty}\frac{(2n-1)^{4k+1}}{e^{(2n-1)\pi}+1}=?$$, $$k\ge1$$ --

$$\int_{-\infty}^{\infty}\frac{\mathrm dt}{(\phi^n t)^2+\pi^2(F_{2n+1}-\phi F_{2n})( At^2+Bt-1)^2}=1$$

B is any values but A must be greater than zero

$$\int_{-\infty}^{\infty}\frac{\mathrm dt}{(\phi^n t)^2+\pi^2(F_{2n+1}-\phi F_{2n})(L_{2n+1}t^2-\phi L_{2n}t-1)^2}=1$$

---

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{L_{n}L_{n+1}}=\frac{\phi+2}{10\phi}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{L_{n+1}L_{n+2}}=\frac{1}{\phi+2}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{L_{n}L_{n+2}}=-\frac{\phi+2}{10\phi}\cdot \frac{1}{\phi^3}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{F_{n+k}F_{n+k+1}}=\frac{1}{\phi^kF_{k}}$$

-

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_n+F_{n+1}}\right)=2\phi^2$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_n+F_{n+1}}\right)^{(-1)^n}=\frac{1}{2}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_n+F_{n+1}}\right)^{(-1)^n+1}=\phi^2$$

-- $$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+1}}\right)^{(-1)^n+1}=\phi^2$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+3}}\right)^{(-1)^n+1}=\left(\frac{2}{\phi}\right)^2$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+4}}\right)^{(-1)^n+1}=\left(\frac{3}{\phi}\right)^2$$

---

$$\prod_{n=2}^{\infty}\frac{F^2_{n+2(k-1)}}{F^2_{n+2(k-1)}+(-1)^n}=\frac{F_{2k}}{F_{2k-1}}\cdot \frac{1}{\phi}$$

$$\prod_{n=2}^{\infty}\frac{F^2_{n+2k-1}}{F^2_{n+2k-1}+(-1)^n}=\frac{F_{2k+1}\cdot F_{2k+2}}{F_{2k-1}\cdot F_{2k}}\cdot \frac{1}{\phi^4}$$

better format

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}\right)=\phi\cdot\frac{F_{2k-1}}{F_{2k}}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}\right)=\phi^4\cdot \frac{F_{2k-1}\cdot F_{2k}}{F_{2k+1}\cdot F_{2k+2}}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+1}}\right)=\frac{\phi^2}{2}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+2}}\right)=\frac{2}{\phi}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+3}}\right)=\frac{3}{\phi^2}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+4}}\right)=\frac{15}{2\phi^4}$$

--

can reverse

$$\prod_{n=2}^{\infty}\left((-1)^n+\frac{1}{F_nF_{n+1}}\right)^{(-1)^n}=2$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+1}}\right)^{(-1)^n}=2$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+3}}\right)^{(-1)^n}=\frac{4}{3}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_nF_{n+4}}\right)^{(-1)^n}=\frac{6}{5}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F^2_{n+1}}\right)^{(-1)^n}=\frac{3}{2}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_{n+2}F_{n+3}}\right)^{(-1)^n}=\frac{10}{9}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_{n+2}F_{n+5}}\right)^{(-1)^n}=\frac{25}{24}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F^2_{n+3}}\right)^{(-1)^n}=\frac{16}{15}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{(-1)^n}{F_{n+4}F_{n+5}}\right)^{(-1)^n}=\frac{65}{64}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_n+F_{n+3}}\right)^{(-1)^n}=1$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_n+F_{n+4}}\right)^{(-1)^n}=\frac{5}{3\phi}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_{n+2}+F_{n+3}}\right)^{(-1)^n}=\frac{3}{4}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_{n+3}+F_{n+2}}\right)^{(-1)^n}=\frac{9}{4}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_{n+2}+F_{n+5}}\right)^{(-1)^n}=1$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_{n+4}+F_{n+5}}\right)^{(-1)^n}=\frac{8}{9}$$

$$\prod_{n=2}^{\infty}\left(1+\frac{1}{(-1)^nF_{n+5}+F_{n+4}}\right)^{(-1)^n}=\frac{4}{3}$$

$$\underset{n=1}{\overset{\infty}}\frac{an}{an-(ak-a+1)}=a$$

$$\underset{n=1}{\overset{\infty}}\frac{n}{2n-1}=\ln{2}$$

$$\underset{n=1}{\overset{\infty}}\frac{n^2}{4n^2-1}=1-\ln{2}$$

$$\underset{n=1}{\overset{\infty}}\frac{\alpha n^x}{\alpha n^x-1}=1$$

$$\underset{n=1}{\overset{\infty}}\frac{n+1}{n-\alpha}=\alpha\frac{1}{\alpha+1}$$

$$\underset{n=1}{\overset{\infty}}\frac{n}{n-\alpha}=\alpha$$

$$\underset{n=1}{\overset{\infty}}\frac{n+\beta}{n-\alpha}=G(\alpha,\beta)$$

$$\underset{n=1}{\overset{\infty}}\frac{a(a+1)}{1}=n$$

$$\underset{n=1}{\overset{\infty}}\frac{a(a+\alpha)}{\alpha}=n$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+2}}\right)=\left(\frac{2}{\phi}\right)^2$$

$$\frac{1-\prod_{n=1}^{\infty}\left(\frac{1}{F_{2^n+1}}\right)}{1-\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^{n+1}}}\right)}=\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^{n}}}\right)$$

a:

$$\sum_{n=1}^{\infty}\left(\frac{1}{F_{(k)2^n}}-\frac{1}{F_{(k+1)2^n}}\right)=\frac{1}{F_{2k+1}-1}$$

$$\prod_{n=1}^{\infty}\left(\frac{1}{F_{(k)2^n+1}}\right)-\prod_{n=1}^{\infty}\left(\frac{1}{F_{(k+1)2^n+1}}\right)=\frac{1}{(F_{2k+1}-1)\phi}$$

b:

$$\sum_{n=1}^{\infty}\left(\frac{1}{F_{(k)2^{n+1}}}-\frac{1}{F_{(k+1)2^{n+1}}}\right)=\frac{1}{F^2_{2k+1}-1}$$

$$\prod_{n=1}^{\infty}\left(\frac{1}{F_{(k)2^{n+1}+1}}\right)-\prod_{n=1}^{\infty}\left(\frac{1}{F_{(k+1)2^{n+1}+1}}\right)=\frac{1}{(F^2_{2k+1}-1)\phi}$$

c:

$$\sum_{n=1}^{\infty}\left(\frac{1}{F_{(k)2^{n+a}}}-\frac{1}{F_{(k+1)2^{n+a}}}\right)=\frac{1}{G(k,a)}$$

$$\prod_{n=1}^{\infty}\left(\frac{1}{F_{(k)2^{n+a}+1}}\right)-\prod_{n=1}^{\infty}\left(\frac{1}{F_{(k+1)2^{n+a}+1}}\right)=\frac{1}{G(k,a)\phi}$$

-

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^{n+1}+1}}\right)+\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}}\right)+\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^{n+1}}}\right)\times\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^n}}\right)=\phi+2$$

$$\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^{n+1}}}\right)+\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^{n+1}+1}}\right)+\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}}\right)\times\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^n}}\right)=\frac{6}{\phi}$$

$$\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}}\right)\times\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^n}}\right)}{\sum_{n=1}^{\infty}\left(\frac{1}{F_{2^{n+1}}}\right)+\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^{n+1}+1}}\right)}=\frac{1}{\phi^4}$$

$$A_4+B_3+C_4D_3=\frac{1}{2}\phi^3$$

--

$$A_{2k}=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{(2k)2^n+1}}\right)$$

$$B_{2k-1}=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{(2k-1)2^n+1}}\right)$$

$$C_{2k}=\sum_{n=1}^{\infty}\left(\frac{1}{F_{(2k)2^n}}\right)$$

$$D_{2k-1}=\sum_{n=1}^{\infty}\left(\frac{1}{F_{(2k-1)2^n}}\right)$$

-

$$\frac{A^2_2-B^2_1}{C_2+D_1}=\frac{5}{\phi^2}$$

$$\frac{A_2+B_1}{C_2D_1}=\phi^3+\phi$$

---

$$\prod_{n=1}^{\infty}\left(3+\frac{L_{n}}{F_{n}}\right)^{(-1)^n}=\phi$$

$$\prod_{n=1}^{\infty}\left(\frac{F_{n}+L_{n}}{2F_{n}+L_{n}}\right)^{(-1)^n}=\frac{2}{\phi}$$

symmetric formulas for $$\phi$$

$$\frac{3}{\sqrt{5}}=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n}+1}\right)}{\sum_{n=1}^{\infty}\frac{1}{F_{2^n}}}$$

$$2\phi=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^{n+1}}+1}\right)}{\sum_{n=1}^{\infty}\frac{1}{F_{2^{n+1}}}}$$

$$?=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n(2k-1)}+1}\right)}{\sum_{n=1}^{\infty}\frac{1}{L_{2^n(2k-1)}}}$$

$$?=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n(2k)}+1}\right)}{\sum_{n=1}^{\infty}\frac{1}{L_{2^n(2k)}}}$$

- $$1=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n(2k-1)}+1}\right)-\frac{1}{\phi}\sum_{n=1}^{\infty}\frac{1}{F_{2^n(2k-1)}}$$

$$1=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n(2k)}+1}\right)-{\frac{1}{\phi}\sum_{n=1}^{\infty}\frac{1}{F_{2^n(2k)}}}$$

$$?={\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n(2k-1)}+1}\right)}-{\frac{1}{\phi}\sum_{n=1}^{\infty}\frac{1}{L_{2^n(2k-1)}}}$$

$$?={\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n(2k)}+1}\right)}-{\frac{1}{\phi}\sum_{n=1}^{\infty}\frac{1}{L_{2^n(2k)}}}$$

--

$$\frac{1}{\phi}=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n(2k)}+1}\right)-\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n(2k-1)}+1}\right)}{\sum_{n=1}^{\infty}\frac{1}{F_{2^n(2k)}}-\sum_{n=1}^{\infty}\frac{1}{F_{2^n(2k-1)}}}$$

--

$$\prod_{n=1}^{\infty}\frac{F^t_{n}-k(-1)^n}{F^t_{(n+2(-1)^n)}-k(-1)^n}=-\frac{k-1}{k+1}\cdot\frac{1}{\phi^{3t}}$$

$$t\ge1$$

--

$$\prod_{n=1}^{\infty}\frac{F_{n}-4(-1)^n}{F_{(n+4(-1)^n)}-4(-1)^n}=-\frac{1}{10^2}\cdot\frac{1}{\phi^{10}}$$

---

$$\prod_{n=2}^{\infty}\frac{F_{n}^2}{F_{n}^2+(-1)^n}=\phi-1$$

$$\prod_{n=2}^{\infty}\frac{F_{n}^2}{F_{n}^2+(-1)^{n+1}}=3-\phi$$

where $$k\ge1$$

$$\phi^{k(2k+1)}\cdot\prod_{n=2}^{\infty}\frac{F_{n}}{F_{[n+2k(-1)^n]}}=g(2k)$$

$$g(2)=1$$, $$g(4)=3$$, $$g(6)=12$$, $$g(8)=\frac{252}{2}$$ and $$g(10)=\frac{2772}{13}$$

$$\phi^{(k-1)(2k-1)}\cdot\prod_{n=2}^{\infty}\frac{F_{n}}{F_{[n+(2k-1)(-1)^n]}}=g(2k-1)$$

$$g(1)=1$$, $$g(3)=3$$, $$g(5)=12$$, $$g(7)=\frac{252}{2}$$ and $$g(9)=\frac{2772}{13}$$

-

$$\prod_{n=2}^{\infty}\frac{F_{n}}{F_{[n+(-1)^n]}}=\frac{1}{\phi}$$

$$\prod_{n=2}^{\infty}\frac{F_{n}}{F_{[n-(-1)^n]}}=1$$

---

$$\prod_{n=1}^{\infty}\frac{F^2_{n}+(-1)^n}{F^2_{[n+(-1)^n]}+(-1)^{n+1}}=\frac{1}{\phi^2}$$

$$\prod_{n=1}^{\infty}\frac{F^2_{n}}{F^2_{[n+(-1)^n]}+(-1)^{n+1}}=\frac{1}{\phi^3}$$

$$\prod_{n=1}^{\infty}\frac{F^2_{n}}{F^2_{[n-(-1)^n]}+(-1)^{n}}=\frac{3}{\phi^4}$$

$$\prod_{n=1}^{\infty}\frac{F^2_{n}}{F^2_{[n+k(-1)^n]}+(-1)^{n+1}}=g(k)$$

-

$$\prod_{n=2}^{\infty}\frac{F_{n}-\pi(-1)^n}{F_{[n+(-1)^n]}+\pi(-1)^{n}}=\frac{1}{\phi}$$

-- telescope:easy

$$\prod_{n=1}^{\infty}\frac{F^t_{n}-k(-1)^n}{F^t_{[n+(-1)^n]}+k(-1)^{n}}=-\left(1+\frac{1}{k}\right)\frac{1}{\phi^t}$$

-

$$\prod_{n=2}^{\infty}\frac{F_{n}-k(-1)^n}{F_{[n-(-1)^n]}+k(-1)^{n}}=-\frac{k-1}{k+1}$$

$$\prod_{n=2}^{\infty}\frac{F_{n}+k(-1)^n}{F_{[n-(-1)^n]}-k(-1)^{n}}=-\frac{k+1}{k-1}$$

--

$$\lim_{k \to \infty}\prod_{j=1}^{k}\left(\prod_{n=2}^{\infty}\frac{F_{n}+(2j-1)(-1)^n}{F_{[n-(-1)^n]}-(2j-1)(-1)^{n}}\right)^{(-1)^j}=\frac{\pi}{2}$$

$$\lim_{k \to \infty}\prod_{j=1}^{k}\prod_{n=2}^{\infty}\frac{F^2_{n}-2(-1)^nF_{n}-16j^2+1}{F^2_{n-(-1)^n}+2(-1)^nF_{n-(-1)^n}-16j^2+1}=\frac{\pi}{2}$$

$$\prod_{n=2}^{\infty}\frac{F^2_{n}-(-1)^nF_{n}-j}{F^2_{n-(-1)^n}+(-1)^nF_{n-(-1)^n}-j}=j$$

$$A=\sqrt[3]{-\frac{1}{2}+\left(\cos\left(\frac{2\pi}{13}\right)+\cos\left(\frac{10\pi}{13}\right)\right)^2}$$

$$B=\sqrt[3]{-\frac{1}{2}+\left(\cos\left(\frac{4\pi}{13}\right)+\cos\left(\frac{6\pi}{13}\right)\right)^2}$$

$$C=\sqrt[3]{-\frac{1}{2}+\left(\cos\left(\frac{8\pi}{13}\right)+\cos\left(\frac{12\pi}{13}\right)\right)^2}$$

$$A+B+C=0$$

-

$$A=\sqrt[3]{-\frac{1}{2}+\left(\frac{1}{2}+\cos\left(\frac{2\pi}{13}\right)+\cos\left(\frac{10\pi}{13}\right)\right)^2}$$

$$B=\sqrt[3]{-\frac{1}{2}+\left(\frac{1}{2}+\cos\left(\frac{4\pi}{13}\right)+\cos\left(\frac{6\pi}{13}\right)\right)^2}$$

$$C=\sqrt[3]{-\frac{1}{2}+\left(\frac{1}{2}+\cos\left(\frac{8\pi}{13}\right)+\cos\left(\frac{12\pi}{13}\right)\right)^2}$$

$$A+B+C=\sqrt[3]{\frac{3\sqrt[3]{13}-2}{4}}$$

---

$$A=\left(-\frac{1}{2}+\left(\cos\left(\frac{2\pi}{13}\right)+\cos\left(\frac{10\pi}{13}\right)\right)^2\right)^{2/3}$$

$$B=\left(-\frac{1}{2}+\left(\cos\left(\frac{4\pi}{13}\right)+\cos\left(\frac{6\pi}{13}\right)\right)^2\right)^{2/3}$$

$$C=\left(-\frac{1}{2}+\left(\cos\left(\frac{8\pi}{13}\right)+\cos\left(\frac{12\pi}{13}\right)\right)^2\right)^{2/3}$$

$$A+B+C=\sqrt[3]{\frac{13}{2}}$$ -- $$k\ge 1$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{(2k-1)2^n+1}}\right)=\frac{L_{2k}}{L_{2k-1}}\cdot \frac{1}{\phi}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{(2k)2^n+1}}\right)=\frac{F_{2k+1}}{F_{2k}}\cdot \frac{1}{\phi}$$

$$\phi=\frac{L_{2k}}{L_{2k-1}}-\sum_{n=0}^{\infty}\frac{1}{F_{2^{n+1}(2k-1)}}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{(2k-1)2^n+1}}\right)=\frac{5F_{2k}}{L_{2k-1}}\cdot \frac{1}{\phi\sqrt{5}}$$

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{(2k)2^n+1}}\right)=\frac{L_{2k+1}}{F_{2k}}\cdot \frac{1}{\phi\sqrt{5}}$$

---

$$\alpha(\beta n+1)^2-\beta^2 n^2+1=(ax+b)(cx+d)$$

where $$\alpha,\beta\ge1$$

--

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{n+1}=\frac{4}{\phi^3}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{(n+1)^2}=2\left[\frac{4}{\phi^3}+8\ln\left(\frac{4}{\phi^3}\right)\right]$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{2n+1}=2\ln{\phi}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{(2n+1)^2}=\frac{\pi^2}{10}$$

--

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{2n-1}=\frac{1}{2}-\phi$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{(2n-1)^2}=\phi-\frac{1}{2}-\frac{1}{2}\ln{\phi}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{1}{(2n-1)^3}=-\frac{\pi^2}{40}-\phi+\frac{1}{2}+\frac{1}{2}\ln{\phi}$$

--

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{n}{(2n-1)^3}=-\frac{\pi^2}{80}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{n}{(2n-1)^2}=-\frac{1}{4}\ln{\phi}$$

- $$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{(2n+1)^2}{(2n-1)^3}=\frac{1}{2}-\phi-\frac{\pi^2}{10}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{4n^2+1}{(2n-1)^3}=\frac{1}{2}-\phi-\frac{\pi^2}{20}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{4^{2n}}\cdot\frac{4n^2-3}{(2n+2)(2n+1)^2}=\frac{2}{\phi^3}-\frac{\pi^2}{5}$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H_{n}}{4^{2n}}=\frac{2}{\sqrt{5}}\ln\left(\frac{\phi^6}{20}\right)$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H_{2n}}{4^{2n}}=\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^6}{25}\right)$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H^{'}_{2n}}{4^{2n}}=\frac{1}{\sqrt{5}}\ln\left(\frac{16}{\phi^6}\right)$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H_{2n+1}}{4^{2n}}=2\ln{\phi}+\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^6}{25}\right)$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H_{2n+2}}{4^{2n}}=\frac{2}{\phi^3}+2\ln{\phi}+\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^6}{25}\right)$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H_{2n+3}}{4^{2n}}=6\phi-7-2\ln{\phi}+\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^6}{25}\right)$$

$$\sum_{n=0}^{\infty}(-1)^n{2n \choose n}\frac{H_{2n+4}}{4^{2n}}=\frac{5}{3}(5-2\phi)-2\ln{\phi}+\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^6}{25}\right)$$

$$A=\sqrt[3]{-1+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}$$ $$B=\sqrt[3]{-1+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}$$ $$C=\sqrt[3]{-1+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}$$

$$A+B+C=0$$

---

$$A=\sqrt[3]{-\frac{1}{2}+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}$$ $$B=\sqrt[3]{-\frac{1}{2}+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}$$ $$C=\sqrt[3]{-\frac{1}{2}+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}$$

$$A+B+C=\sqrt[3]{\frac{3}{2}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)}$$ --- $$A=\sqrt[3]{-\frac{1}{3}+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}$$ $$B=\sqrt[3]{-\frac{1}{3}+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}$$ $$C=\sqrt[3]{-\frac{1}{3}+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}$$

$$A+B+C=\sqrt[3]{3\left(1+\sqrt[3]{7}\right)}$$

$$A=\sqrt[3]{-\frac{2}{3}+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}$$ $$B=\sqrt[3]{-\frac{2}{3}+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}$$ $$C=\sqrt[3]{-\frac{2}{3}+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}$$

$$A+B+C=\sqrt[3]{3\left(2-\sqrt[3]{7}\right)}$$

$$A=\left({-1+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$B=\left({-1+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$C=\left({-1+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}\right)^{2/3}$$

$$A+B+C=2\sqrt[3]{7}$$

--

$$A=\left({-\frac{1}{2}+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$B=\left({-\frac{1}{2}+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$C=\left({-\frac{1}{2}+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}\right)^{2/3}$$

$$A+B+C=\sqrt[3]{\frac{89+15\sqrt[3]{7}+9\sqrt[3]{49}}{4}}$$

$$A=\left({-\frac{1}{3}+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$B=\left({-\frac{1}{3}+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$C=\left({-\frac{1}{3}+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}\right)^{2/3}$$

$$A+B+C=\sqrt[3]{23+2\sqrt[3]{7}+3\sqrt[3]{49}}$$

---

$$A=\left({-\frac{2}{3}+\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$B=\left({-\frac{2}{3}+\left(1+2\cos\left(\frac{4\pi}{7}\right)\right)^2}\right)^{2/3}$$ $$C=\left({-\frac{2}{3}+\left(1+2\cos\left(\frac{8\pi}{7}\right)\right)^2}\right)^{2/3}$$

$$A+B+C=\sqrt[3]{20+2\sqrt[3]{7}+3\sqrt[3]{49}}$$

$$\sum_{n=0}^{\infty}\frac{F_{2n}}{p^n}=\frac{p}{p^2-3p+1}$$ p=3...

$$\sum_{n=0}^{\infty}(-1)^n\frac{F_{[n+(2a-1)(-1)^n]}}{2^n}=2F_{(a+1)}$$ a=1,2,3,4,..

$$\sum_{n=0}^{\infty}(-1)^n\frac{F_{[(2a-1)+n+(-1)^n]}}{2^n}=2F_{(a+1)}$$ a=1,2,3,4,...

$$\sum_{n=0}^{\infty}(-1)^n\frac{F_{[a+n+(-1)^n]}}{2^n}=2F_{(a)}$$ a=1,2,3,4,...

$$\sum_{n=0}^{\infty}\frac{1}{F_{[1+n+(-1)^n]}}=\sum_{n=0}^{\infty}\frac{1}{F_{[1+n+3(-1)^n]}}=\sum_{n=1}^{\infty}\frac{1}{F_{n}}=W$$

$$\sum_{n=0}^{\infty}\frac{1}{F_{[a+n+(-1)^n]}}=W-a+1$$, a:={1,2 and 3}

$$\lim_{N\to \infty} N-\sum_{n=0}^{N}\frac{F_{n+2}^2}{F_{n+1}F_{n+3}}=-\frac{2}{\phi^2}$$

$$\lim_{N\to \infty} N-\sum_{n=0}^{N}\frac{F_{n+2}^3F_{n+4}}{F_{n+1}F_{n+3}^3}=-\frac{2}{\phi^2}$$

$$\sum_{n=0}^{\infty}\left[\left(\frac{F_{n+2}^2}{F_{n+1}F_{n+3}}\right)^x-\left(\frac{F_{n+3}^2}{F_{n+2}F_{n+4}}\right)^x\right] =\frac{1-2^x}{2^x}$$

$$\sum_{n=0}^{\infty}\left[\left(\frac{F_{n+2}^3F_{n+4}}{F_{n+1}F_{n+3}^3}\right)^x-\left(\frac{F_{n+3}^3F_{n+5}}{F_{n+2}F_{n+4}^3}\right)^x\right]=\frac{3^x-8^3}{8^x}=\frac{3^x-2^{3x}}{2^{3x}}$$

$$\sum_{n=0}^{\infty}\left[\frac{F_{n+k+1}}{F_{n+k}}-\frac{F_{n+k+2}}{F_{n+k+1}}\right]=\frac{(-1)^k}{\phi^kF_{k}}$$

--

$$A_n=\cos^n\left(\frac{2\pi}{13}\right)+\cos^n\left(\frac{10\pi}{13}\right)$$

$$B_n=\cos^n\left(\frac{4\pi}{13}\right)+\cos^n\left(\frac{6\pi}{13}\right)$$

$$C_n=\cos^n\left(\frac{8\pi}{13}\right)+\cos^n\left(\frac{12\pi}{13}\right)$$

$$A_1^{2/3}+B_1^{2/3}+C_1^{2/3}=\sqrt[3]{\frac{15}{4}+\frac{3}{2}\sqrt[3]{13}+\frac{3}{4}\sqrt[3]{13^2}}$$

$$(-1+A_2)^{1/3}+(-1+B_2)^{1/3}+(-1+C_2)^{1/3}=\sqrt[3]{\frac{3\sqrt[3]{13}-7}{4}}$$

$$(-1+A_2)^{2/3}+(-1+B_2)^{2/3}+(-1+C_2)^{2/3}=\sqrt[3]{\frac{15}{16}+\frac{3}{8}\sqrt[3]{13}+\frac{3}{16}\sqrt[3]{13^2}}$$

$$(-1/2+A_1)^{2/3}+(-1/2+B_1)^{2/3}+(-1/2+C_1)^{2/3}=\sqrt[3]{5+\frac{3}{2}\sqrt[3]{13}-\frac{3}{4}\sqrt[3]{13^2}}$$

---

$$A=\cos\left(\frac{\pi}{19}\right)+\cos\left(\frac{7\pi}{19}\right)+\cos\left(\frac{11\pi}{19}\right)=x_1+y_1+z_1$$

$$B=\cos\left(\frac{3\pi}{19}\right)+\cos\left(\frac{5\pi}{19}\right)+\cos\left(\frac{17\pi}{19}\right)x_2+y_2+z_2$$

$$C=\cos\left(\frac{9\pi}{19}\right)+\cos\left(\frac{13\pi}{19}\right)+\cos\left(\frac{15\pi}{19}\right)x_3+y_3+z_3$$

$$D=\cos^2\left(\frac{\pi}{19}\right)+\cos^2\left(\frac{7\pi}{19}\right)+\cos^2\left(\frac{11\pi}{19}\right)$$

$$E=\cos^2\left(\frac{3\pi}{19}\right)+\cos^2\left(\frac{5\pi}{19}\right)+\cos^2\left(\frac{17\pi}{19}\right)$$

$$F=\cos^2\left(\frac{9\pi}{19}\right)+\cos^2\left(\frac{13\pi}{19}\right)+\cos^2\left(\frac{15\pi}{19}\right)$$

$$\left(-\frac{1}{2}+A\right)^{2/3}+\left(-\frac{1}{2}+B\right)^{2/3}+\left(-\frac{1}{2}+C\right)^{2/3}=\sqrt[3]{\frac{27}{4}+\frac{3}{2}\sqrt[3]{19}+\frac{3}{4}\sqrt[3]{19^2}}$$

$$\left(-1+A\right)^{2/3}+\left(-1+B\right)^{2/3}+\left(-1+C\right)^{2/3}=\sqrt[3]{5+\frac{3}{2}\sqrt[3]{19}+\frac{3}{4}\sqrt[3]{19^2}}$$

$$\left(-1+D\right)^{2/3}+\left(-1+E\right)^{2/3}+\left(-1+F\right)^{2/3}=\sqrt[3]{\frac{27}{16}+\frac{3}{8}\sqrt[3]{19}+\frac{3}{16}\sqrt[3]{19^2}}$$

$$\left(-1+D\right)^{1/3}+\left(-1+E\right)^{1/3}+\left(-1+F\right)^{1/3}=\sqrt[3]{-\frac{1}{4}+\frac{3}{4}\sqrt[3]{19}}$$

$$\left(-1+A\right)^{1/3}+\left(-1+B\right)^{1/3}+\left(-1+C\right)^{1/3}=\sqrt[3]{\frac{1}{2}+\frac{3}{2}\sqrt[3]{19}}$$

$$\left(x_1y_1z_1\right)^{1/3}+\left(x_2y_2z_2\right)^{1/3}+\left(x_3y_3z_3\right)^{1/3}=\sqrt[3]{\frac{1-3\sqrt[3]{19}}{8}}$$

$$\left(x_1y_1z_1\right)^{2/3}+\left(x_2y_2z_2\right)^{2/3}+\left(x_3y_3z_3\right)^{2/3}=\sqrt[3]{\frac{27+6\sqrt[3]{19}+3\sqrt[3]{19^2}}{64}}$$

$$\left(x_1y_1z_1\right)\left(x_2y_2z_2\right)\left(x_3y_3z_3\right)=\frac{1}{2}$$

--

$$\int_{0}^{1}\frac{x^2(1-3x^2)}{1+x^2}\mathrm dx=3-\pi$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2-a)(x^2+3)}{(1+x^2)^3}\mathrm dx=\frac{a-1}{4}(10-3\pi)$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2-1)(ax^2+b)}{(1+x^2)^3}\mathrm dx=\frac{3a-b}{4}(22-7\pi)$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2-5)(x^2+5)}{(1+x^2)^5}\mathrm dx=\frac{1}{12}$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2-5)(2x^2-11)}{(1+x^2)^5}\mathrm dx=\frac{\pi}{16}$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2+5)(5x^2-11)}{(1+x^2)^5}\mathrm dx=\frac{1}{8}(22-7\pi)$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2+5)(5x^2-11)^3}{(1+x^2)^5}\mathrm dx=203-59\pi$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(x^2+5)[ (n+1)x^2+n-15]}{(1+x^2)^3}\mathrm dx=n(4\pi-13)+3$$

$$\int_{0}^{1}\frac{x^2(1-3x^2)(ax^2+a+4)(5x^2-11)}{(1+x^2)^3}\mathrm dx=79a-7^2-\pi(5^2a-4^2)$$

-

Arithmetic-Fibonacci sequences

Where $$a_1,a_2,a_3,a_4,...a_n$$ are the consecutive terms in the arithmetic progression, let $$\lim{n \to \infty}$$ then

an example:

5, 9, 13, 17, ...ntn=4n+1

$$\frac{\pi}{4}=1+\sum_{n=1}^{\infty}(-1)^n\frac{(4n+9)^2-(4n+5)^2}{(8n+9)^2-5^2}$$

$$\frac{\pi}{4}=1+\sum_{n=1}^{\infty}(-1)^n\frac{a_{n+2}^2-a_{n+1}^2}{a_{2n+2}^2-a_1^2}$$

Where $$F_k,F_{k+1},F_{k+2},F_{k+3},...F_{k+n}$$ are the consecutive terms in the Fibonacci sequence, let $$\lim{n \to \infty}$$ then

$$\frac{\pi}{4}=\sum_{n=1}^{\infty} \arctan\left[\frac{F_{k+n+1}^2+F_{k+n}^2}{F_{k+2n+1}^2+F_k^2}\right]$$

---

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(\phi^n t)^2+(F_{2n+1}-\phi F_{2n})(e^{\gamma}t^2+\zeta(3)t-\pi)^2}=1$$

$$\int_{-\infty}^{+\infty}\frac{t^2}{(\phi^n t)^2+(F_{2n+1}-\phi F_{2n})(\pi t^2+\zeta(3)t-e^{\gamma})^2}\mathrm dt=1$$

$$\int_{-\infty}^{+\infty}\frac{t^2}{(\phi^n t)^2+\pi^2(F_{2n+1}-\phi F_{2n})(t^2+\zeta(3)t-e^{\gamma})^2}\mathrm dt=1$$

$$\int_{-\infty}^{+\infty}\frac{t^2}{(\phi^n t)^2+\pi(F_{2n+1}-\phi F_{2n})(\sqrt{\pi}t^2+\zeta(3)t-e^{\gamma})^2}\mathrm dt=1$$

-- $$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{ax^2+b(x^2-x-c)^2}=\frac{\pi}{c\sqrt{ab}}$$

$$\int_{-\infty}^{+\infty}\frac{x^2\mathrm dx}{ax^2+b(cx^2+x-1)^2}=\frac{\pi}{c\sqrt{ab}}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{x^2+(e^{\gamma}x^2+\phi x-1)^2}=1$$

$$\int_{-\infty}^{+\infty}\frac{x^2\mathrm dx}{x^2+\pi^2(x^2+\phi x-e^{\gamma})^2}=1$$

$$\int_{-\infty}^{+\infty}\frac{x^2\mathrm dx}{x^2+\pi(\sqrt{\pi}x^2+\phi x-e^{\gamma})^2}=1$$

$$\int_{-\infty}^{+\infty}\frac{x^2\mathrm dx}{d(x^2+x-1)^2+a(x^3+bx^2-cx-1)^2}=\frac{\pi}{\sqrt{a\cdot d}}\cdot \frac{c}{c^2+c(b-2)-(b-2)^2}$$

$$\int_{-\infty}^{+\infty}\frac{x^2\mathrm dx}{(x^2+x-1)^2+(x^3+3x^2-x-1)^2}=\pi$$

$$\int_{-\infty}^{+\infty}\frac{x^2\mathrm dx}{(x^2+x-1)^2+(x^3+2x^2-x-1)^2}=\pi$$

$$\sum_{n=-\infty}^{+\infty}\frac{1}{n^2+(n+1)^2+2x(x+1)}=F(x)$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{x^2+(x+1)^2+2n(n+1)}=\frac{\pi}{2n+1}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(x^2+(x+1)^2+2n(n+1))^2}=\frac{\pi}{(2n+1)^3}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{x^2+(2x-n)^2}=\frac{\pi}{n}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{x^2+(x+n)^2}=\frac{\pi}{n}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{x^2+(x^2-x-n)^2}=\frac{\pi}{n}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{x^2+(x^2+x+1)^2}=\pi\cdot \sqrt{\frac{\phi}{5}}$$

$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(x^2+x)^2+(x^2+x+1)^2}=2\pi\cdot \sqrt{\frac{1}{5\phi}}$$

-

$$\left(1+\cos\left(\frac{2\pi}{5}\right)\right)^{1/4}+\left(1+\cos\left(\frac{4\pi}{5}\right)\right)^{1/4}=\left(\frac{1}{2}\phi^6\right)^{1/4}$$

$$\sqrt[3]{2a+2\cos\left(\frac{2\pi}{5}\right)}-2\sqrt[3]{2a+2\cos\left(\frac{4\pi}{5}\right)}+\sqrt[3]{2a+2\cos\left(\frac{8\pi}{5}\right)}=-\sqrt[3]{2(2a-\phi)}+\sqrt[3]{2(2a+\phi^{-1})}$$

$$\sqrt[3]{2a+2\cos\left(\frac{2\pi}{5}\right)}+2\sqrt[3]{2a+2\cos\left(\frac{4\pi}{5}\right)}+\sqrt[3]{2a+2\cos\left(\frac{8\pi}{5}\right)}=\sqrt[3]{2(2a-\phi)}+\sqrt[3]{2(2a+\phi^{-1})}$$

$$\sqrt[3]{4a+4\cos\left(\frac{2\pi}{5}\right)}+\sqrt[3]{4a+4\cos\left(\frac{4\pi}{5}\right)}+\sqrt[3]{4a+4\cos\left(\frac{8\pi}{5}\right)}=\sqrt[3]{2(2a-\phi)}+2\sqrt[3]{2(2a+\phi^{-1})}$$

$$\sqrt[3]{4a+4\cos\left(\frac{2\pi}{5}\right)}-\sqrt[3]{4a+4\cos\left(\frac{4\pi}{5}\right)}+\sqrt[3]{4a+4\cos\left(\frac{8\pi}{5}\right)}=-\sqrt[3]{2(2a-\phi)}+2\sqrt[3]{2(2a+\phi^{-1})}$$

$$\sqrt[3]{1+\cos\left(\frac{2\pi}{9}\right)}+\sqrt[3]{1+\cos\left(\frac{4\pi}{9}\right)}+\sqrt[3]{1+\cos\left(\frac{8\pi}{9}\right)}=\sqrt[3]{6+\frac{9}{2}\sqrt[3]{3}+3\sqrt[3]{3}}$$

$$\left({1+\cos\left(\frac{2\pi}{9}\right)}\right)^{2/3}+\left({1+\cos\left(\frac{4\pi}{9}\right)}\right)^{2/3}+\left({1+\cos\left(\frac{8\pi}{9}\right)}\right)^{2/3}=\sqrt[3]{6+\frac{9}{2}\sqrt[3]{3}+\frac{15}{4}\sqrt[3]{9}}$$

$$\left({2+2\cos\left(\frac{2\pi}{7}\right)}\right)^{2/3}+\left({2+2\cos\left(\frac{4\pi}{7}\right)}\right)^{2/3}+\left({2+2\cos\left(\frac{8\pi}{7}\right)}\right)^{2/3}=\sqrt[3]{19+9\sqrt[3]{7}+6\sqrt[3]{49}}$$

$$\left({2+2\cos\left(\frac{2\pi}{9}\right)}\right)^{2/3}+\left({2+2\cos\left(\frac{4\pi}{9}\right)}\right)^{2/3}+\left({2+2\cos\left(\frac{8\pi}{9}\right)}\right)^{2/3}=\sqrt[3]{24+18\sqrt[3]{3}+15\sqrt[3]{9}}$$

$$\int_{-\infty}^{+\infty}\frac{x(x+a)(x+b)(x+c)}{(x^3+2x^2-x-1)^2+(x^2+x-1)^2}=[a^2+b^2+c^2-a-b-c-3(d^2-1)]\pi$$

where a,b and c are consecutive arithmetic terms and d is the common difference -

$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(\alpha x^2-x-\frac{(\alpha-1)^2}{2(2\alpha-1)}\right)^2}=\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(\alpha x^2-\alpha x-\frac{\alpha-1}{2}\right)^2}=\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x-2)^4}{(2x-1)^4+\left(x^2-x+1\right)^4}=\frac{22}{5}\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x+1)^4}{(2x-1)^4+\left(x^2-x+1\right)^4}=\frac{22}{5}\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x-1/2)^4}{(2x-1)^4+\left(x^2-x+1\right)^4}=\frac{1}{20}\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x+1)^2}{(x^2-x-1)^2+\left((2\alpha+1)x^2+x-\alpha\right)^2}=\frac{\pi}{\alpha+1}$$

$$\int_{-\infty}^{+\infty}\frac{(x+1)^2}{(x^2-x-1)^2+\left((2\alpha-1)x^2+x-2\alpha+1\right)^2}=\frac{\pi}{\alpha}$$

$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(x^3+2x^2-x-1)^2+(5x^2+x-1)^2}=70\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(x^3+2x^2-x-1)^2+(x^2+x-1)^2}=2\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x+1)^2}{(x^3+2x^2-x-1)^2+(5x^2+x-1)^2}=6\pi$$

$$\int_{-\infty}^{+\infty}\frac{(x+1)^2}{(x^3+2x^2-x-1)^2+(x^2+x-1)^2}=2\pi$$

-

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{\beta}}{(2n-1)^2}-\frac{(n-\alpha)^{\beta}}{(n+1)^2}\right]=8(-\alpha-1)^{\beta}$$

valid $$...,-3,-2,-1,\le \alpha\le 0,1,2,3,...$$

and valid $$...,-3,-2,-1,\le \beta\le 0,1,2,3,...$$

$$\alpha\ge0$$ and $$\beta=\frac{1}{2},\frac{1}{3}...$$ looking for closed form.

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{1/2}}{(2n-1)^2}-\frac{(n-\alpha)^{1/2}}{(n+1)^2}\right]=i8(\alpha+1)^{1/2}$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{1/3}}{(2n-1)^2}-\frac{(n-\alpha)^{1/3}}{(n+1)^2}\right]=4(1+i\sqrt{3})(\alpha+1)^{1/3}$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{1/4}}{(2n-1)^2}-\frac{(n-\alpha)^{1/4}}{(n+1)^2}\right]=4\sqrt{2}(1+i)(\alpha+1)^{1/4}$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{1/6}}{(2n-1)^2}-\frac{(n-\alpha)^{1/6}}{(n+1)^2}\right]=4(\sqrt{3}+i)(\alpha+1)^{1/6}$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{1/5}}{(2n-1)^2}-\frac{(n-\alpha)^{1/5}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\phi^2+i\sqrt{\phi\sqrt{5}}\right)(\alpha+1)^{1/5}$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{8(n-\alpha-1)^{1/10}}{(2n-1)^2}-\frac{(n-\alpha)^{1/10}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\sqrt{\phi^3\sqrt{5}}+i\right)(\alpha+1)^{1/10}$$

-

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(n+1)x^{n+1}}=\frac{2}{\sqrt{x}}\tan^{-1}\left(\frac{1}{\sqrt{x}}\right)-\ln\left(\frac{x+1}{x}\right)$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(n+1)}\eta(n+1)=\ln\frac{2}{\pi}+2\sum_{n=0}^{\infty}\frac{(-1)^n}{\sqrt{n+1}}\tan^{-1}\left(\frac{1}{\sqrt{n+1}}\right)$$

--

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2{2n\choose n}}=\frac{\pi^2}{6}-3\ln^2{\phi}$$

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2{2n\choose n}}=\frac{8G-\pi\ln{(2+\sqrt{3})}}{3}$$

--

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)^2\left[5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}\right]}=F_{k+1}$$ --

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)^3}=\frac{1}{5}\left[1+\frac{4}{\sqrt{5}}\ln{(\phi)}\right]$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)^4}=\frac{1}{25}\left[6+\frac{4}{\sqrt{5}}\ln{(\phi)}\right]$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)(n+x)}=2(x-1)\ln^2{(\phi)}+\frac{2}{\sqrt{5}}\ln{\phi}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)^2(5n+3)}=1$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)^3(5n+4)}=1$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{x}\right)^n(n+1)}=2x\ln^2\left[\frac{1+\sqrt{4x+1}}{2\sqrt{x}}\right]$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{x(1+x)}\right)^n(n+1)}=\frac{x(1+x)}{2}\ln^2{\left[\frac{1+x}{x}\right]}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n(n+1)}=2\ln^2{(\phi)}$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^n(n+1)}=\ln^2{(2)}$$ --

$$\sum_{n=0}^{\infty}\frac{(-1)^n}=\frac{2}{5}\zeta(3)$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^n}=\frac{\zeta(3)}{2}-\frac{\ln^3{(2)}}{3}$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n n}=2\ln^2{(\phi)}-\frac{2}{5}\zeta(3)$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^n n}=-\frac{\zeta(3)}{2}+\frac{\ln^3{(2)}}{3}+\ln^2{(2)}$$

---

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{n}{(n+1)^2}-\frac{8}{2n-1}\right]=-\frac{8\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{n}{(n+1)^2}+\frac{8}{(2n-1)^2}\right]=+\frac{8\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{n}{(n+1)^2}-\frac{4}{2n-1}+\frac{4 }{(2n-1)^2}\right]=8$$

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{n-\alpha}{(n+1)^2}-\frac{8(n-\alpha-1)}{(2n-1)^2}\right]=8(\alpha+1)$$

-

$$\sum_{n=0}^{\infty}\frac{2^n}\frac{1}{2n+1}\left[\frac{1}{(n+1)^2}-\frac{1}{(n+2)^2}\right]=\frac{\pi^2+4\pi-16}{8}$$

---

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{16^n}\frac{1}{(2n-1)(2n-3)\cdots[2n-2k-1]}=(-1)^{k+1}\frac{2}{\pi}\frac{(2k)!!}{(2k+1)!!^2}$$, $$k\ge0$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{16^n}\frac{1}{[(2n-1)(2n-3)\cdots[2n-2k-1]^2}=\frac{2}{\pi}\frac{(2k)!!^2}{(2k+1)!!^3}\frac{2^{k+1}}{k!}$$, $$k\ge0$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{16^n}\frac{n}{(2n-1)(2n-3)\cdots[2n-2k-1]}=(-1)^{k}\frac{1}{2k+1}\frac{1}{\pi}\frac{(2k-2)!!}{(2k-1)!!^2}$$, $$k\ge1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{16^n}\frac{n^2}{(2n-1)(2n-3)(2n-5)\cdots[2n-2k-1]}=?$$, $$k\ge2$$

$$\sum_{n=1}^{\infty}\frac{24^{2n}}(16n+1)=\frac{3\sqrt{3}}{\sqrt{\pi^3}}\Gamma^2\left(\frac{3}{4}\right)= \sqrt{\left(\frac{3}{\pi}\right)^3}\Gamma^2\left(\frac{3}{4}\right)=$$

$$\sum_{n=1}^{\infty}\frac{24^{2n}}n(4n+1)=\frac{3\sqrt{3}}{32\sqrt{\pi^3}}\Gamma^2\left(\frac{3}{4}\right)=$$

$$\sum_{n=1}^{\infty}\frac{24^{2n}}\frac{1}{(n+1)(n+2)\cdots (n+k)}=?$$

$$\sum_{n=1}^{\infty}\frac{(2n+1)}\frac{1}{16^n}=\sqrt{2}$$

$$\sum_{n=1}^{\infty}\frac{(2n+1)(2n+3)}\frac{(-1)^n}{48^n}=4-\frac{8}{3^{3/4}}$$

$$\sum_{n=1}^{\infty}\frac{(2n+1)(2n+2)}\frac{(-1)^n}{48^n}=4-\frac{8}{3^{3/4}}$$

$$\frac{3}{4}\sum_{n=1}^{\infty}\frac{(2n+1)(2n+3)}\frac{(-1)^n}{48^n}=x-\sqrt[4]{x(1+x)^3}\cdot\cos\left[\frac{3}{2}\tan^{-1}\left(\frac{1}{\sqrt{x}}\right)\right]$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{4n}}\frac{H_{n+1}}{(2n-1)^2}=\frac{140-144\ln{2}-9\pi}{9\pi}$$

-

$$\sum_{n=1}^{\infty}\frac{1}{{2n \choose n }(2n+1)(n+1)^3}=\frac{17\pi^4}{1620}-1$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{35n-37}{(2n-1)(n-1)^2}+\frac{35n+37}{(2n+1)(n+1)^2}\right]=8\ln{(2)}$$

--

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n-1}{(2n-1)^2}+\frac{n-1}{2n-1}\right]=\beta(2)-\beta(1)$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n-1}{(2n-1)^3}+\frac{n-1}{(2n-1)^2}\right]=\beta(3)-\beta(2)$$

stage 1:

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{1}{(2n-1)^s(n-1)}+\frac{1}{(2n+1)^s(n+1)}\right]=\frac{1}{2\cdot3^s}$$

stage 2:

(+4)

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-58}{(2n-1)^3(n-1)}+\frac{2(n^2+n)-58}{(2n+1)^3(n+1)}\right]=-\frac{\pi^3}{32}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-490}{(2n-1)^5(n-1)}+\frac{2(n^2+n)-490}{(2n+1)^5(n+1)}\right]=-\frac{5\pi^5}{1536}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-4378}{(2n-1)^7(n-1)}+\frac{2(n^2+n)-4378}{(2n+1)^7(n+1)}\right]=-\frac{61\pi^7}{184320}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-39370}{(2n-1)^7(n-1)}+\frac{2(n^2+n)-39370}{(2n+1)^7(n+1)}\right]=-\frac{277\pi^9}{82557536}$$

Final formulae:

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-2(3^s+2)}{(2n-1)^s(n-1)}+\frac{2(n^2+n)-2(3^s+2)}{(2n+1)^s(n+1)}\right]=F(s)$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-2(2\cdot 3^s+1)}{(2n-1)^s(n-1)}+\frac{n^2+n-2(2\cdot3^s+1)}{(2n+1)^s(n+1)}\right]=F(s)$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{\alpha(n^2-n)-2[(\alpha-\beta)\cdot 3^s+\beta]}{(2n-1)^s(n-1)}+\frac{\beta(n^2+n)-2[(\alpha-\beta)\cdot3^s+\beta]}{(2n+1)^s(n+1)}\right]=F(s)$$

stage 3:

(-2)

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4}{(2n-1)(n-1)}+\frac{n^2+n+4}{(2n+1)(n+1)}\right]=\frac{\pi}{4}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{52}{(2n-1)^3(n-1)}+\frac{n^2+n+52}{(2n+1)^3(n+1)}\right]=\frac{\pi^3}{32}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{482}{(2n-1)^5(n-1)}+\frac{n^2+n+482}{(2n+1)^5(n+1)}\right]=\frac{5\pi^5}{1536}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]=\frac{61\pi^7}{184320}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{39374}{(2n-1)^7(n-1)}+\frac{n^2+n+39374}{(2n+1)^7(n+1)}\right]=\frac{277\pi^9}{82557536}$$

Final formulae:

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{2(3^s-1)}{(2n-1)^s(n-1)}+\frac{n^2+n+2(3^s-1)}{(2n+1)^s(n+1)}\right]=F(s)$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n^2-n-2\cdot 3^s}{(2n-1)^s(n-1)}-\frac{2\cdot(3^s-1)}{(2n+1)^s(n+1)}\right]=G(s)$$

---

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{6n}\left[5\cdot\frac{4n^2-1}{n^2-1}-\left(\frac{2n-1}{n-1}\right)^2-\left(\frac{2n+1}{n+1}\right)^2\right]=\ln{(2)}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{1}{(2n-1)^s(n-1)}+\frac{1}{(2n+1)^s(n+1)}\right]=\frac{1}{(3)^s\cdot2}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{19}{(2n-1)(n-1)}+\frac{18}{(2n+1)(n+1)}\right]=\pi$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{a}{(2n-1)(n-1)}+\frac{b}{(2n+1)(n+1)}\right]=\frac{6\pi(a-b)-18a+19b}{6}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{a}{(2n-1)(n-1)^2}+\frac{b}{(2n+1)(n+1)^2}\right]=\frac{(a+b)[\pi^2-24\pi+49]+35a+24(b-a)\ln{(2)}}{12}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n+1)}\left[\frac{a}{(2n-1)(n-1)}+\frac{b}{(2n+1)(n+1)}\right]=\frac{b\pi^2+8\pi(a-3b)-8(a-3b)\ln{(2)-19a+49b}}{12}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n+1)}\left[\frac{4n+7}{(2n-1)(n-1)}+\frac{4n+5}{(2n+1)(n+1)}\right]=\frac{\pi^2}{12}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n+2)}\left[\frac{14n+23}{(2n-1)(n-1)}+\frac{12n+23}{(2n+1)(n+1)}\right]=\frac{2\pi}{3}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n+3)}\left[\frac{337n+843}{(2n-1)(n-1)}+\frac{321n+843}{(2n+1)(n+1)}\right]=16\pi$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n+3)}\left[\frac{7}{(2n-1)(n-1)}+\frac{5}{(2n+1)(n+1)}\right]=\frac{11}{48}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{a}{(2n-1)^3(n-1)}+\frac{b}{(2n+1)^3(n+1)}\right]=\frac{27\pi^3(a-b)+216\pi(a-b)-1512a+1516b}{216}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[2\cdot\frac{103n^2-103n+28}{(2n-1)^3(n-1)}+\frac{215n^2+215n+56}{(2n+1)^3(n+1)}\right]=\frac{9\pi^3}{32}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{413n^2-413n+110}{(2n-1)^3(n-1)}+\frac{431n^2+431n+110}{(2n+1)^3(n+1)}\right]=\frac{9\pi^3}{16}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n^2-n-2}{(2n-1)^3(n-1)}+\frac{n^2+n-2}{(2n+1)^3(n+1)}\right]=0$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{52}{(2n-1)^3(n-1)}+\frac{n^2+n+52}{(2n+1)^3(n+1)}\right]=\frac{\pi^3}{32}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n^2-n-54}{(2n-1)^3(n-1)}-\frac{54}{(2n+1)^3(n+1)}\right]=-\frac{\pi^3}{32}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{a}{(2n-1)^5(n-1)}+\frac{b}{(2n+1)^5(n+1)}\right]=F(x)$$

$$f(x)=\frac{405\pi^5(a-b)+3888\pi^3(a-b)+31104\pi(a-b)-342144a+342208b}{31104}$$

-

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4P(n)}{(2n-1)^5(n-1)}+\frac{Q(n)}{(2n+1)^5(n+1)}\right]=\frac{135\pi^5}{512}$$

Where $$P(n)=3806n^4-7612n^3+1497n^2+2309n-805$$

and $$Q(n)=15224n^4+30448n^3+6069n^2-9155n-3220$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^5(n-1)}+\frac{Q(n)}{(2n+1)^5(n+1)}\right]=\frac{135\pi^5}{256}$$

Where $$P(n)=15580n^4-31160n^3+23127n^2-7547n+994$$

and $$Q(n)=15580n^4+31160n^3+23289n^2+7709n+994$$

---

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^5(n-1)}+\frac{Q(n)}{(2n+1)^5(n+1)}\right]=\frac{135\pi^5}{256}$$

Where $$P(n)=15584n^4-31168n^3+23133n^2-7549n+974$$

and $$Q(n)=15584n^4+31168n^3+23295n^2+7711n+974$$ -

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^5(n-1)}+\frac{Q(n)}{(2n+1)^5(n+1)}\right]=0$$

Where $$P(n)=2n^4-4n^3+3n^2-n-10$$

and $$Q(n)=2n^4+4n^3+3n^2+n-10$$ ---

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{3(n^2-n)-974}{(2n-1)^5(n-1)}+\frac{n^2+n-974}{(2n+1)^5(n+1)}\right]=-\frac{5\pi^5}{768}$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^5(n-1)}+\frac{Q(n)}{(2n+1)^5(n+1)}\right]=\frac{135\pi^5}{256}$$

Where $$P(n)=15572n^4-31144n^3+23115n^2-7543n+1034$$

and $$Q(n)=15572n^4+31144n^3+23277n^2+7705n+1034$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^7(n-1)}+\frac{Q(n)}{(2n+1)^7(n+1)}\right]=\frac{4941\pi^7}{20480}$$

Where $$P(n)=280208(n^2-n)^3+210156(n^2-n)^2+50352(n^2-n)+4196$$

and $$Q(n)=280208(n^2+n)^3+210156(n^2+n)^2+51081(n^2+n)+4196$$

-

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^7(n-1)}+\frac{Q(n)}{(2n+1)^7(n+1)}\right]=\frac{4941\pi^7}{10240}$$

Where $$P(n)=560432(n^2-n)^3+420324(n^2-n)^2+100707(n^2-n)+8210$$

and $$Q(n)=560432(n^2+n)^3+420324(n^2+n)^2+102165(n^2+n)+8210$$

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{P(n)}{(2n-1)^7(n-1)}+\frac{Q(n)}{(2n+1)^7(n+1)}\right]=0$$

Where $$P(n)=16(n^2-n)^3+12(n^2-n)^2+3(n^2-n)-182$$

and $$Q(n)=16(n^2+n)^3+12(n^2+n)^2+3(n^2+n)-182$$

--

-

$$\prod_{n=1}^{\infty}\left[\frac{n+1}{n}\right]^{(-1)^n}=\frac{2}{\pi}$$

$$\prod_{n=1}^{\infty}\left[\frac{n+1}{n}\cdot \frac{n+3}{n+2}\right]^{(-1)^n}=\frac{2^4}{3\pi^2}$$

$$\prod_{n=1}^{\infty}\left[\frac{n+1}{n}\cdot \frac{n+3}{n+2}\cdot \frac{n+5}{n+4}\right]^{(-1)^n}=\frac{2^{11}}{135\pi^3}$$

-

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}H_{n+1}=\frac{32\pi+\left(1-\frac{4\ln{2}}{\pi}\right)\Gamma^4\left(\frac{1}{4}\right)}{4\sqrt{\pi}\Gamma^2 \left(\frac{1}{4}\right)}-1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{1}{n+1}\left[1+\frac{8}{2n-1}+\frac{6}{(2n-1)^2}\right]=1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{n}{n+1}\left[1+\frac{8}{2n-1}+\frac{6}{(2n-1)^2}\right]=\frac{8\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{n}{(n+1)(n+2)(2n-1)^2}=\frac{4}{25}\frac{\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{n}{(2n-1)^2}\frac{1}{(n+1)(n+2)\cdots(n+2k)}=\left[\frac{2^k}{(4k+1)!!}\right]^2\frac{\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$, $$k\ge1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{n}{(2n-1)^2}\frac{1}{(n+1)(n+2)\cdots(n+2k-2)(n+2k-1)}=\left[\frac{\Gamma^2\left(\frac{1}{4}\right)}{(4k-1)!!}\right]^2\frac{1}{\sqrt{\pi^3}}$$, $$k\ge2$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}=\frac{\Gamma^2\left(\frac{1}{4}\right)}{2\sqrt{\pi^3}}-1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{1}{n+1}=\frac{8\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}-1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{1}{(n+1)^2}=\frac{32\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}+\frac{2\Gamma^2\left(\frac{1}{4}\right)}{\sqrt{\pi^3}}-9$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[4+\frac{4}{n+1}-\frac{1}{(n+1)^2}\right]=1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{1}{2n-1}=1-\frac{2\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}-\frac{\Gamma^2\left(\frac{1}{4}\right)}{4\sqrt{\pi^3}}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\frac{1}{(2n-1)^2}=-1+\frac{4\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}+\frac{\Gamma^2\left(\frac{1}{4}\right)}{4\sqrt{\pi^3}}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{(n+1)^2}-\frac{8}{(2n-1)^2}\right]=1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{(n+1)^2}-\frac{8}{(2n-1)^2}\right]n=8-\frac{32\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}-\frac{2\Gamma^2\left(\frac{1}{4}\right)}{\sqrt{\pi^3}}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{n+1}-\frac{8n}{(2n-1)^2}\right]=-1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{n+1}-\frac{8n}{(2n-1)^2}\right]n=\sqrt{\frac{2}{\pi}}\frac{\Gamma\left(-\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{n+1}-\frac{8n^2}{(2n-1)^2}\right]=-1+\frac{8\sqrt{\pi}}{\Gamma^2\left(\frac{1}{4}\right)}-\frac{\Gamma^2\left(\frac{1}{4}\right)}{2\sqrt{\pi^3}}$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{n+1}+2\cdot\frac{2n+1}{2n-1}\right]=1$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{n+1}+\frac{4}{2n-1}+\frac{16n^2}{(2n-1)^2}\right]=3$$

-- $$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{2n-1}+2n\cdot\frac{2n+1}{(2n-1)^2}\right]=0$$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{2n-1}+2n\cdot\frac{2n+1}{(2n-1)^2}\right]=1$$

$$\sum_{n=2}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{2n-1}+2n\cdot\frac{2n+1}{(2n-1)^2}\right]=\frac{1}{8}$$

$$\sum_{n=3}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{2n-1}+2n\cdot\frac{2n+1}{(2n-1)^2}\right]=\frac{9}{2^8}$$

$$\sum_{n=4}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{2n-1}+2n\cdot\frac{2n+1}{(2n-1)^2}\right]=\frac{25}{2^{11}}$$

$$\sum_{n=s}^{\infty}\frac{{2n \choose n}^2}{2^{5n}}\left[\frac{1}{2n-1}+2n\cdot\frac{2n+1}{(2n-1)^2}\right]=F(s)$$

---

$$\sum_{n=0}^{\infty}\frac{15n^2+7n}{{3n \choose n}2^n}=\frac{\pi}{2}+\frac{18}{5}$$

$$\sum_{n=0}^{\infty}\frac{5n^2-6n+1}{{3n \choose n}2^n}=\sum_{n=0}^{\infty}\frac{(5n-1)(n-1)}{{3n \choose n}2^n}=\frac{6}{5}$$

--

$$\sum_{n=0}^{\infty}\frac\frac{1}{x^n}\left[\frac{1}{(n+1)^k}+\frac{1}{(n+1)^{k+1}}\right]=2x\left[2Li_k(x^{-1})-Li_{k+1}(x^{-1})\right]$$

$$\sum_{n=0}^{\infty}\frac\frac{1}{x^n}\left[\frac{1}{(n+1)^k}-\frac{1}{(n+1)^{k+2}}\right]=2x\left[2Li_k(x^{-1})-3Li_{k+1}(x^{-1})+Li_{k+2}(x^{-1})\right]$$

$$\sum_{n=0}^{\infty}\frac\frac{1}{x^n}\left[\frac{1}{(n+1)^2}+\frac{(-1)^{y+1}}{(n+1)^{2+y}}\right]=2x\left[2\sum_{j=2}^{y}(-1)^jli_j(1/x)+\sum_{k=3}^{y}(-1)^k li_k(1/x)\right]$$

$$\sum_{n=0}^{\infty}\frac\frac{1}{x^n}\left[\frac{1}{(n+1)^s}+\frac{(-1)^{y+1}}{(n+1)^{s+y}}\right]=2x(-1)^{y+1}\left[2\sum_{j=s}^{y}(-1)^jLi_j(1/x)+\sum_{k=s+1}^{y}(-1)^{k} Li_k(1/x)\right]$$

--- $$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{1}{(2n+1)(n+1)}=\frac{\pi^2}{4}$$

$$\sum_{n=1}^{\infty}\frac{4^n}\cdot \frac{1}{(2n+1)(n+1)(n+2)}=\frac{\pi^2-4}{8}$$

$$\sum_{n=2}^{\infty}\frac{4^n}\cdot \frac{1}{(2n+1)(n+2)}=\frac{\pi^2+4}{8}$$

$$\sum_{n=3}^{\infty}\frac{4^n}\cdot \frac{1}{(2n+1)(n+2)}=\frac{\pi^2+4}{8}$$

$$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{1}{2n+1}\left[\frac{a}{n+1}+\frac{b}{n+2}\right]=\frac{2a\pi^2+b(\pi^2+4)}{8}$$

$$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{1}{(2n+1)(n+1)}\left[\frac{a}{n+1}+\frac{b}{n+2}\right]=-\frac{7a}{4}\zeta(3)+\frac{1}{2}a\pi^2\ln{(2)}+\frac{1}{8}b(\pi^2-4)$$

$$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{n}{(2n+1)(n+1)}\left[\frac{a}{n+1}+\frac{b}{n+2}\right]=\frac{a}{4}[7\zeta(3)-\pi^2(\ln{(4)-1})]+b$$

$$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{n}{(2n+1)(n+1)(n+2)}=1$$

$$\sum_{n=0}^{\infty}\frac{2^{2n}}\frac{1}{(n+1)^2(2n+1)}=\frac{1}{2}\pi^2\ln{(2)}-\frac{7}{4}\zeta(3)$$

$$\sum_{n=0}^{\infty}\frac{2^{n}}\frac{1}{(n+1)^2(2n+1)}=\frac{1}{8}\pi^2\ln{(2)}-\frac{35}{16}\zeta(3)+\pi G$$

$$\sum_{n=0}^{\infty}\frac{2^{n}-2^{2n-2}}\frac{1}{(n+1)^2(2n+1)}=\pi G-\frac{7}{4}\zeta(3)$$

$$\sum_{n=0}^{\infty}\frac{2^{n}-5\times 2^{2n-2}}\frac{1}{(n+1)^2(2n+1)}=\pi G-\frac{1}{2}\pi^2\ln{(2)}$$

---

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+3}\left[\frac{1}{n+1}+\frac{2}{n+2}+\frac{1}{n+3}\right]=\frac{(3\pi+2)(5\pi-14)}{16}$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(2n+1)(2n+3)}=\left[\frac{a}{n+1}+\frac{b}{n+2}\right]=-\frac{a(\pi^2+8\pi-32)-b(\pi^2+4\pi-24)}{8}$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(2n+1)(2n+3)}=\left[\frac{a}{n+1}+\frac{2b}{n+2}+\frac{c}{n+3}\right]=a\left(\frac{\pi^2}{8}+\pi-4\right)-\frac{b}{4}\left(\pi^2+4\pi-24\right)-\frac{c}{16}(\pi^2-12)$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(n+3)}=\left[\frac{a}{n+1}+\frac{b}{n+2}\right]=-\frac{1}{32}(\pi-2)[a(\pi-14)+2b(\pi-6)]$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(n+1)(n+2)}=\left[\frac{1}{n+1}+\frac{1}{n+2}\right]=\frac{\pi^2+4\pi-16}{8}$$

--

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(n+1)}=\left[\frac{1}{n+1}+\frac{1}{n+2}\right]=\pi G-\frac{35}{16}\zeta(3)-1+\frac{\pi}{2}+\frac{1}{8}\pi^2\ln{(2)}$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(n+1)}=\left[\frac{1}{n+1}+\frac{2}{n+2}+\frac{1}{n+3}\right]=\pi G-\frac{35}{16}\zeta(3)-\frac{23}{8}+\frac{3\pi}{2}-\frac{\pi^2}{32}+\frac{1}{8}\pi^2\ln{(2)}$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{(2n+1)(n+1)}=\left[\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}\right]=\pi G-\frac{35}{16}\zeta(3)+\frac{9}{8}-\frac{\pi}{2}-\frac{\pi^2}{32}+\frac{1}{8}\pi^2\ln{(2)}$$

---

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}\right]=\frac{\pi^2}{8}$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}-\frac{1}{n+2}\right]=\frac{1}{2}[\pi-2]$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}\right]=\frac{1}{16}[\pi^2-4]$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}-\frac{3}{n+2}+\frac{3}{n+3}-\frac{1}{n+4}\right]=\frac{1}{9}[3\pi-7]$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}-\frac{4}{n+2}+\frac{6}{n+3}-\frac{4}{n+4}+\frac{1}{n+5}\right]=\frac{1}{64}[3\pi^2-16]$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}-\frac{5}{n+2}+\frac{10}{n+3}-\frac{10}{n+4}+\frac{5}{n+5}-\frac{1}{n+6}\right]=\frac{1}{225}[60\pi-149]$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{2}{n+2}-\frac{5}{n+3}+\frac{4}{n+4}-\frac{1}{n+5}\right]=\frac{\pi^2}{64}$$

$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{4}{n+1}-\frac{40}{n+2}+\frac{54}{n+3}-\frac{18}{n+4}\right]=-\pi$$

---

1.$$\sum_{n=0}^{\infty}\frac{H_{n+k}}\cdot \frac{2^n}{2n+1}=F(k)$$

$$F(0)=2G-\frac{\pi}{2}\ln{(2)}$$

$$F(1)=2G-\frac{\pi}{2}\ln{(2)}+\frac{\pi^2}{8}$$

$$F(2)=2G-\frac{\pi}{2}\ln{(2)}+\frac{\pi^2}{4}-\frac{\pi}{2}+1$$

$$F(3)=2G-\frac{\pi}{2}\ln{(2)}+\frac{7\pi^2}{16}-\frac{3\pi}{2}+\frac{11}{4}$$

$$F(4)=2G-\frac{\pi}{2}\ln{(2)}+\frac{3\pi^2}{4}-\frac{10\pi}{3}+\frac{52}{9}$$

2.$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=F(n)$$

3.$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}+\frac{1}{n+2}\right]=\frac{4-2\pi+\pi^2}{4}$$

3(a).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{n+1}+\frac{1}{n+2}\right]^2=\frac{32\pi G+4\pi^2\ln{(2)}-2\pi^2-70\zeta(3)+8\pi}{16}$$

3(b).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[aH_{n}+\frac{b}{n+1}+\frac{c}{n+2}\right]=\frac{16aG-4a\pi\ln{(2)}+b\pi^2+(8-4\pi+\pi^2)c}{8}$$

3(c).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{a}{(n+1)^2}+\frac{b}{n+2}\right]=\frac{a}{16}\left[16\pi G-35\zeta(3)+2\pi^2\ln{(2)}\right]+\frac{b}{8}(8-4\pi+\pi^2)$$

3(d).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{a}{n+1}+\frac{b}{n+2}\right]=\frac{a\pi^2+b(8-4\pi+\pi^2)}{8}$$

3(e).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{a}{n+1}-\frac{2b}{n+2}+\frac{c}{n+3}\right]=\frac{1}{8}a\pi^2+b\left(-2+\pi-\frac{\pi^2}{4}\right)+\frac{c}{3}\left(\frac{21}{4}-3\pi+\frac{9\pi^2}{16}\right)$$

4.$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{(n+1)^2}-\frac{1}{(n+2)^2}\right]=\frac{-16+4\pi+\pi^2}{8}$$

4(a).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{a}{(n+1)^2}+\frac{b}{(n+2)^2}\right]=\frac{a+b}{16}\left[16\pi G+2\pi^2\ln{(2)}-35\zeta(3)\right]+\frac{b}{8}\left[4-2\pi-\pi^2\right]$$

4(b).$$\sum_{n=0}^{\infty}\frac{2^n}\cdot \frac{1}{2n+1}=\left[\frac{1}{(n+1)^2}-\frac{3}{(n+2)^2}+\frac{3}{(n+3)^2}-\frac{1}{(n+4)^2}\right]=\frac{-812+192\pi+45\pi^2}{432}$$

1.$$\sum_{n=0}^{\infty}\frac{(-1)^n}{{2n\choose n}2^{n}}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)=12+22\ln^2{(2)}-\frac{92}{3}\ln{(2)}$$

2.$$\sum_{n=0}^{\infty}\frac{1}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)=\frac{-54+26\sqrt{3}\pi-7\pi^2}{9}$$

3.$$\sum_{n=0}^{\infty}\frac{n^3}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)=\frac{4482-1588\sqrt{3}\pi+441\pi^2}{81}$$

4.$$\sum_{n=0}^{\infty}\frac{1}\left(\frac{1}{(2n+1)(n+3)}+\frac{1}{(2n+3)(n+2)}\right)=\frac{33}{2}-\pi\sqrt{27}$$

1.$$\sum_{n=0}^{\infty}\frac{1}{{2n\choose n}x^{2n}}\left(\frac{1}{2n+1}+\frac{1}{2n+2}\right)=2x^2\sin^{-1}\left(\frac{1}{2x}\right)\left[\frac{4}{\sqrt{4x^2-1}}-\sin^{-1}\left(\frac{1}{2x}\right)\right]$$

2.$$\sum_{n=0}^{\infty}\frac{1}{{2n\choose n}x^{2n}}\left(\frac{1}{2n+1}-\frac{2}{2n+2}+\frac{1}{2x+3}\right)=4x^2\left[-2+2\sqrt{4x^2-1}\sin^{-1}\left(\frac{1}{2x}\right)+\sin^{-1}\left(\frac{1}{2x}\right)\right]$$

--

1.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)^2(2n+3)}=G-\frac{1}{2}$$

2.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)^2(2n+2k+1)}=\frac{G}{k}-F(k)$$, $$k\ge1$$

3.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{n^2}=\frac{1}{2}(\pi-2)(\pi+2)$$

4.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(n+1)^2}=\frac{7}{4}\zeta(3)+\frac{\pi^2}{2}[1-ln{(2)}]-1$$

5.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(n+1)^2}=\frac{1}{4}(\pi-2)(\pi+2)$$

6.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(n+2)^2}=\frac{7}{8}\zeta(3)+\frac{\pi^2}{4}[1-ln{(2)}]-\frac{3}{4}$$

7.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(n+1)^2}=\frac{1}{4}(\pi^2+4)$$

8.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(n+2)^2}=\frac{\pi^2}{16}$$

9.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)^2}=2G$$

10.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)^2(n+1)}=4G-\frac{\pi^2}{4}$$

1.$$\sum_{n=0}^{\infty}\frac{1}\cdot \frac{1}{(2n+1)(n+1)x^{2n}}=4x^2\sin^{-1}\left(\frac{1}{2x}\right)^2$$

2.$$\sum_{n=0}^{\infty}\frac{1}\cdot \frac{1}{(2n+1)x^{2n}}=\frac{4x^2}{\sqrt{4x^2-1}}\sin^{-1}\left(\frac{1}{2x}\right)$$

3.$$\sum_{n=0}^{\infty}\frac{2}\cdot \frac{1}{(2n+1)(n+1)(n+2)}=-\frac{2}{9}\pi^2+\frac{4}{\sqrt{3}}\pi-4$$

4.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)(2n+3)}=\frac{2G+1}{4}$$

5.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)(2n+2)}=G$$

6.$$\sum_{n=1}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)(2n+2)(2n+3)}=\frac{2G-1}{4}$$

7.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)(2n+2)(2n+3)}=\frac{2G-1}{8}$$

8.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)(2n+2)}=\frac{G}{2}$$

9.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{1}{(2n+1)(2n+3)}=\frac{2G+1}{8}$$

10.$$\sum_{n=0}^{\infty}\frac{2^{2n}}\cdot \frac{n}{(2n+1)^2(2n+3)(n+1)}=\frac{\pi^2}{4}-G-\frac{3}{2}$$

--- Ramanujan's formulas

1(a).$$\sum_{n=0}^{\infty}(-1)^n\frac{{2n \choose n}^3}{2^{6n}}=\frac{\sqrt{2}-1}{8\sqrt{2} \pi^2}\cdot \frac{\Gamma^4\left(\frac{1}{8}\right)}{\Gamma^2\left(\frac{1}{4}\right)} $$

2(a).$$\sum_{n=0}^{\infty}(-1)^n\frac{{2n \choose n}^3}{2^{6n}}n=\frac{1}{2\pi}-\frac{\sqrt{2}-1}{32\sqrt{2} \pi^2}\cdot \frac{\Gamma^4\left(\frac{1}{8}\right)}{\Gamma^2\left(\frac{1}{4}\right)}$$

3(a).$$\sum_{n=0}^{\infty}(-1)^n\frac{{2n \choose n}^3}{2^{6n}}(4n+1)=\frac{2}{\pi}$$

1(b).$$\sum_{n=0}^{\infty}(1)^n\frac{{2n \choose n}^3}{2^{8n}}=?$$

2(b).$$\sum_{n=0}^{\infty}(1)^n\frac{{2n \choose n}^3}{2^{8n}}n=\frac{2}{3\pi}-\frac{\Gamma^6\left(\frac{1}{3}\right)}{8\times 2^{2/3}\sqrt{3}\pi^4}$$

3(b).$$\sum_{n=0}^{\infty}(1)^n\frac{{2n \choose n}^3}{2^{8n}}(6n+1)=\frac{4}{\pi}$$

1(c).$$\sum_{n=0}^{\infty}(1)^n\frac{{2n \choose n}^3}{2^{12n}}=?$$

2(c).$$\sum_{n=0}^{\infty}(1)^n\frac{{2n \choose n}^3}{2^{12n}}n=\frac{8}{21\pi}-\frac{5}{168}\frac{\Gamma^2\left(\frac{1}{7}\right)\Gamma^2\left(\frac{2}{7}\right)\Gamma^2\left(\frac{4}{7}\right)}{\sqrt{7}\pi^4}$$

3(c).$$\sum_{n=0}^{\infty}(1)^n\frac{{2n \choose n}^3}{2^{12n}}(42n+5)=\frac{16}{\pi}$$

conjectures:

Where A, B, C and D are integers.

1.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{2^{6kn}}n=\frac{A}{B\pi}+\frac{C}{D}f(k)$$ then

2.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{2^{6kn}}(2Bn+C)=\frac{2A}{\pi}$$ then

1(a).$$\sum_{n=0}^{\infty}(-1)^n\frac{{2n \choose n}^3}{2^{(6k+2)n}}n=\frac{A}{B\pi^2}+\frac{C}{D}f(k)$$ then

2(a).$$\sum_{n=0}^{\infty}(-1)^n\frac{{2n \choose n}^3}{2^{(6k+2)n}}(2Bn+C)=\frac{2A}{\pi}$$ then

-

1.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}H_{n,2}}{4^n}\cdot \frac{1}{(2n-1)}=4(1-\ln{2})$$

-

1.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1}{(n+1)(n+3)(2n-1)^2}=\frac{x}{\pi}$$

2.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1}{(n+1)(n+3)(n+5)(2n-1)^2}=\frac{y}{\pi}$$

3.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{(4n-1)(2kn+1)}{(2n-1)^3}=\frac{2(k+1)(2G+1)}{\pi}$$ -

1.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}=\frac{\pi}{\Gamma^4\left(\frac{3}{4}\right)}$$

2.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{n}{n+1}=\frac{16\pi}{\Gamma^4\left(\frac{1}{4}\right)}$$

3.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{1}{n+1}=\pi\left(\frac{1}{\Gamma^4\left(\frac{3}{4}\right)}-\frac{16}{\Gamma^4\left(\frac{1}{4}\right)}\right)$$

3(a).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{(n-1)(4n-1)}{(2n-1)^2}=\pi\left(\frac{1}{\Gamma^4\left(\frac{3}{4}\right)}-\frac{8}{\Gamma^4\left(\frac{1}{4}\right)}\right)$$

4.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{n[An(2n-1)^2+B(n+1)}{(n+1)(2n-1)^2}=\frac{8\pi(2A+B)}{\Gamma^4\left(\frac{1}{4}\right)}$$

5.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{An(2n-1)^3+B(n+1)}{(n+1)(2n-1)^3}=\frac{16\pi(A-3B)}{\Gamma^4\left(\frac{1}{4}\right)}$$

---

1.$$\sum_{n=0}^{\infty}\frac{16^n}\cdot \frac{n^g}{(2n+1)(2n-1)^2}=F(g)$$

1(a).$$\sum_{n=0}^{\infty}\frac{16^n}\cdot \frac{1+12n^2}{(2n+1)(2n-1)^2}=\frac{\pi}{2}$$

1(b).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1+12n^2}{(2n+1)(2n-1)^2}=\frac{4}{\pi}[G+1]$$

--

1.$$\sum_{n=0}^{\infty}\frac{16^n}\cdot \frac{n}{(2n-1)^2}=\frac{\pi}{24}$$

-

1.$$\sum_{n=0}^{\infty}\frac{16^n}\cdot \frac{1}{(n+1)(2n-1)^2}=\frac{16+\pi}{18}$$

1(a).$$\sum_{n=0}^{\infty}\frac{16^n}\cdot \frac{5n-4n^3}{(n+1)(2n+1)(2n-1)^2}=\frac{16-5\pi}{18}$$

1(b).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1-6n}{((2n+1)(2n-1)^2}=\frac{4G-2}{\pi}$$

1(c).$$\sum_{n=0}^{\infty}\frac{16^n}\cdot \frac{1-6n}{(2n+1)(2n-1)^2}=\frac{\pi}{4}$$

1(d).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1-6n}{(2n+1)(2n-1)^3}=\frac{1-6G}{\pi}$$

1(e).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{n^2(1-6n)}{(2n+1)(2n-1)^3}=-\frac{1+6G}{4\pi}$$

1(f).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{n(1-6n)}{(2n+1)(2n-1)^3}=-\frac{1+2G}{2\pi}$$

1(g).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1-6n}{(2n-1)^3}=-\frac{8G}{\pi}$$

1(1g).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{n(1-6n)}{(2n-1)^3}=-\frac{1+4G}{\pi}$$

1(2g).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1+6n}{(2n-1)^3}=-\frac{4(4G-3)}{\pi}$$

1(3g).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{n(1+6n)}{(2n-1)^3}=\frac{(8G-1)}{\pi}$$

1(4g).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{n^2}{(2n-1)^3}=\frac{G}{\pi}$$

1(h).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{n}{(2n-1)^2}=\frac{1}{\pi}$$

1(2h).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1}{(2n-1)}=-\frac{2}{\pi}$$

1(i).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{n}{(2n-1)^2}=\frac{8\pi}{\Gamma^4\left(\frac{1}{4}\right)}$$

1(j).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{1}{(2n-1)^3}=-\frac{48\pi}{\Gamma^4\left(\frac{1}{4}\right)}$$

1(k).$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^3}{64^n}\cdot \frac{12n^2-6n-1}{(2n-1)^3}=\frac{96\pi}{\Gamma^4\left(\frac{1}{4}\right)}$$

2.$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{16^n}\cdot \frac{1}{(n+1)(2n-1)^2}=\frac{32}{9\pi}$$

--

1.$$\sum_{n=0}^{\infty}\frac{(16x^2)^n}\cdot \frac{1}{(2n+1)(2n+2)(2n+3)}=\frac{3x\sqrt{4x^2-1}+2x\sin^{-1}(1/2x)+4x^3\sin^{-1}(1/2x)-8x^2}{2}$$

2.$$\sum_{n=0}^{\infty}\frac{(16)^n}\cdot \frac{n}{(2n+1)(2n+2)}=4-2\sqrt{3}-\frac{\pi}{6}$$

3.$$\sum_{n=0}^{\infty}\frac{(8)^n}\cdot \frac{n}{(2n+1)(2n+2)}=2-\sqrt{2}-\frac{\pi}{4\sqrt{2}}$$

3(a).$$\sum_{n=0}^{\infty}\frac{(8)^n}\cdot \frac{1}{(2n+1)(2n+2)}=-2+\sqrt{2}+\frac{\pi}{2\sqrt{2}}$$

3(b).$$\sum_{n=0}^{\infty}\frac{(8)^n}\cdot \frac{1}{(2n+1)^2(2n+2)}=\frac{G}{\sqrt{2}}+2-\sqrt{2}+\frac{\pi}{4\sqrt{2}}[\ln{2}-2]$$

3(c).$$\sum_{n=0}^{\infty}\frac{(8)^n}\cdot \frac{1}{(2n+1)^2}=\frac{G}{\sqrt{2}}+\frac{\pi\ln{2}}{4\sqrt{2}}$$

3(d).$$\sum_{n=0}^{\infty}\frac{(8)^n}\cdot \frac{1}{(2n+1)^3}=\sqrt{2}G+4-2\sqrt{2}+\frac{\pi}{2\sqrt{2}}[\ln{2}-2]$$

3(e).$$\sum_{n=0}^{\infty}\frac{8^n}\cdot \frac{2n+3}{(n+1)(2n+1)^3}=\sqrt{2}\left[+\frac{8}{3}G-\frac{128}{27}+\frac{\pi}{4}(6\ln{2}-10)\right]+\frac{248}{27}$$

3(f).$$\sum_{n=0}^{\infty}\frac{8^n}\cdot \frac{1}{(n+1)(2n+1)^3}=\sqrt{2}\left[+\frac{4}{3}G-\frac{88}{27}+\frac{\pi}{9}(\ln{8}-8)\right]+\frac{184}{27}$$

3(g).$$\sum_{n=0}^{\infty}\frac{8^n}\cdot \frac{1}{(2n+1)^3}=\sqrt{2}\left[+\frac{2}{3}G-\frac{20}{27}+\frac{\pi}{9}(\ln{8}-2)\right]+\frac{32}{27}$$

3(h).$$\sum_{n=0}^{\infty}\frac{8^n}\cdot \frac{1}{(2n+1)^3}=\sqrt{2}\left[2G-4+\frac{\pi}{2}(\ln{2}-2)\right]+8$$

3(i).$$\sum_{n=0}^{\infty}\frac{8^n}\cdot \frac{1}{(2n+1)^3}=\sqrt{2}\left[+\frac{8}{3}G-\frac{128}{27}+\frac{\pi}{9}(6\ln{2}-10)\right]+\frac{248}{27}$$

4.$$\sum_{n=0}^{\infty}\frac{(8)^n}\cdot \frac{n}{(2n+1)(2n+2)}=1-\frac{\pi}{4}$$

1.$$\sum_{n=0}^{\infty}\left(\frac{1}{16x}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{2n+1}=\sqrt{\frac{2}{1+\sqrt{\frac{x-1}{2}}}}$$

1(a).$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{(2n+1)(2n+3)}=\frac{4}{15}\cdot \frac{1}{\sin(\pi/4)}$$

1(b).$$\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{(2n+1)(2n+3)}=\frac{2}{15}\cdot \frac{1}{\sin(\pi/8)}$$

1(c).$$\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{(2n+1)(2n+3)(2n+5)}=\frac{8}{63}\cdot \frac{1}{\sin(\pi/4)}\cdot\sin(\pi/8)$$

1(d).$$\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{(2n+1)(2n+3)(2n+5)(2n+7)}=\frac{8}{2145}\cdot \frac{1}{\sin(\pi/8)}$$

1(e).$$\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}=\frac{32}{16065}\cdot \frac{1}{\sin(\pi/4)}\cdot \sin(\pi/8)$$

2.$$\sum_{n=0}^{\infty}\left(\frac{1}{16x}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{1}{(2n+1)(2n+2)\cdots(2n+k)}=F(k,x)$$

3.$$\sum_{n=0}^{\infty}\left(\frac{1}{16x}\right)^{n}\left[{4n \choose 2n}\right]\cdot \frac{n}{(2n+1)(2n+2)\cdots(2n+k)}=F(k,x)$$

1.$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\left[2{2n \choose n+1}+{2n \choose n+2}\right]\cdot \frac{1}{(2n+1)(2n+2)}=\frac{18\sqrt{3}-28-\pi}{3}$$

2.$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\left[{2n \choose n+3}+3{2n+1 \choose n+2}\right]\cdot \frac{1}{(2n+1)(2n+2)}=\frac{642\sqrt{3}-1096-5\pi}{15}$$

3.$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\left[8{2n \choose n+1}+7{2n \choose n+2}+2{2n \choose n+3}\right]\cdot \frac{1}{(2n+1)(2n+2)}=\frac{1374\sqrt{3}-2332}{15}-\pi$$

4.$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\left[4{2n \choose n+1}+5{2n \choose n+2}+2{2n \choose n+3}\right]\cdot \frac{1}{(2n+1)(2n+2)}=\frac{398\sqrt{3}-684}{5}-\frac{\pi}{3}$$

5.$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\left[{2n+2 \choose n}\right]\cdot \frac{6n+11}{(2n+1)(2n+2)(2n+3)}=\frac{8}{3}[7-2\pi]$$

--

1.$$\sum_{n=0}^{\infty}\left(\frac{1}{2x}\right)^{2n}{2n \choose n}\cdot \frac{1}{(2n+1)(2n+2)}=x\sqrt{x^2-1}-x^2+\sin^{-1}\left(\frac{1}{x}\right)$$

2.$$\sum_{n=0}^{\infty}\left(\frac{1}{2x}\right)^{2n}{2n \choose n+1}\cdot \frac{1}{(2n+1)(2n+2)}=F(x)$$

2(a).$$\sum_{n=0}^{\infty}\left(\frac{1}{8}\right)^{n}{2n \choose n+1}\cdot \frac{1}{(2n+1)(2n+2)}=6-\left(3+\frac{\pi}{4}\right)\sqrt{2}-4\ln(4-2\sqrt{2})$$

2(b).$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}{2n \choose n+1}\cdot \frac{1}{(2n+1)(2n+2)}=12-6\sqrt{3}-\frac{\pi}{3}-8\ln(8-4\sqrt{3})$$

3.$$\sum_{n=0}^{\infty}\left(\frac{1}{2x}\right)^{2n}{2n+2 \choose n}\cdot \frac{1}{(2n+1)(2n+2)}=F(x)$$

3(a).$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}{2n+2 \choose n}\cdot \frac{1}{(2n+1)(2n+3)}=\frac{4}{3}[2\pi+15\sqrt{3}-32]$$

3(b).$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}{2n+2 \choose n}\cdot \frac{1}{(2n+1)(2n+2)(2n+3)}=-\frac{4}{3}[2\pi+9\sqrt{3}-22]$$

4.$$\sum_{n=0}^{\infty}\left(\frac{1}{2x}\right)^{2n}{2n \choose n+2}\cdot \frac{1}{(2n+1)(2n+2)}=F(x)$$

4(a).$$\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}{2n \choose n+2}\cdot \frac{1}{(2n+1)(2n+2)}=\frac{\pi}{2}-\frac{13}{3}+4\ln{2}$$

4(b).$$\sum_{n=0}^{\infty}\left(\frac{1}{8}\right)^{n}{2n \choose n+2}\cdot \frac{1}{(2n+1)(2n+2)}=\frac{19\sqrt{2}-34}{3}+\frac{\pi}{4}\sqrt{2}+8\ln{({4-2\sqrt{2}})}$$

4(c).$$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}{2n \choose n+2}\cdot \frac{1}{(2n+1)(2n+2)}=\frac{54\sqrt{3}-100}{3}+\frac{\pi}{3}+16\ln{({8-4\sqrt{3}})}$$

1.$$\sum_{n=1}^{\infty}\frac{3^n}\cdot\frac{1}{(n+1)^2}=\frac{4\pi}{27}-1$$

2.$$\sum_{n=1}^{\infty}\frac{3^n}\cdot\frac{1}{(n+1)(n+2)(n+3)(n+4)}=\frac{4\pi}{729}-\frac{1}{24}$$

1.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot n=\frac{\pi}{2}+2$$

2.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot n^2=2\pi+6$$

3.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot H_n=\frac{4G+\pi(1-\ln{2})}{2}$$

3.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot H_n=\frac{4G+\pi(1-\ln{2})}{4}$$

1.$$\sum_{n=1}^{\infty}\frac{2^n}=\frac{\pi}{2}$$

2.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{n}=1$$

3.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{n^2}=\frac{\pi(\pi+4)}{8}-2$$

4.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{n+1}=\frac{\pi}{2}-1$$

5.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{n+2}=-\frac{1}{8}(\pi-6)(\pi-2)$$

6.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{(n+1)(n+2)}=\frac{1}{8}(\pi-2)^2$$

7.$$\sum_{n=1}^{\infty}\frac{2^n}\cdot\frac{1}{(n+1)(n+2)(n+4)}=\frac{1}{288}(8-3\pi)^2$$

1. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{1^n}{(2n+3)(2n+5)}=\frac{817}{45}-\frac{10\pi}{\sqrt{3}}$$

2. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{2^n}{(2n+3)(2n+5)}=\frac{167}{45}-\frac{7\pi}{\sqrt{6}}$$

3. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{3^n}{(2n+3)(2n+5)}=\frac{19}{15}-\frac{52\pi}{81\sqrt{3}}$$

4. $$\sum_{n=1}^{\infty}\frac{k^n}\cdot \frac{n^g}{(2n+1)(2n+3)(2n+5)}=F(k,g)$$

5. $$\sum_{n=1}^{\infty}\frac{k^n}\cdot \frac{n^g}{(2n+3)(2n+5)(2n+7)}=F(k,g)$$

1. $$\sum_{n=1}^{\infty}\frac{(-8)^n}H^{'}_{2n}=\frac{\sqrt{6}}{6}\ln\left[40-16\sqrt{6}\right]$$

2. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{x^n}{2n+1}=\frac{4\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}}-1$$

3. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{(-x)^n}{2n+1}=\frac{4\sinh^{-1}\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4+x)}}-1$$

4. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{x^n}{(2n+1)(2n+2)}=\frac{2\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)^2}{x}-\frac{1}{2}$$

5. $$\sum_{n=1}^{\infty}\frac{1}\cdot \frac{(-x)^n}{(2n+1)(2n+2)}=\frac{2\sinh^{-1}\left(\frac{\sqrt{x}}{2}\right)^2}{x}-\frac{1}{2}$$

6. $$\sum_{n=1}^{\infty}\frac{n}\cdot \frac{x^n}{(2n+1)(2n+2)}=-\frac{2\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)^2}{x}+\frac{2\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}}$$

7. $$\sum_{n=1}^{\infty}\frac{n^2}\cdot \frac{3^n}{(2n+1)(2n+2)}=1+\frac{2\pi^2}{27}$$

8. $$\sum_{n=1}^{\infty}\frac{n^2}\cdot \frac{2^n}{(2n+1)(2n+2)}=\frac{8-4\pi+\pi^2}{16}$$

9. $$\sum_{n=1}^{\infty}\frac{n^2}\cdot \frac{1^n}{(2n+1)(2n+2)}=\frac{18-8\sqrt{3}\pi+3\pi^2}{54}$$

10. $$\sum_{n=1}^{\infty}\frac{n}\cdot \frac{1^n}{(2n+1)(2n+2)}=\frac{(2\sqrt{3}-\pi)\pi}{18}$$

11. $$\sum_{n=1}^{\infty}\frac{n}\cdot \frac{2^n}{(2n+1)(2n+2)}=-\frac{(\pi-4)\pi}{16}$$

12. $$\sum_{n=1}^{\infty}\frac{n}\cdot \frac{3^n}{(2n+1)(2n+2)}=\frac{2\pi(3\sqrt{3}-\pi)}{27}$$

1. $$\sum_{n=1}^{\infty}\frac{2^n}{n^2(n+1)(n+2){2n \choose n}}=\frac{1}{32}(\pi-2)(\pi+2)$$

2. $$\sum_{n=1}^{\infty}\frac{2^n}{n^2(n+2){2n \choose n}}=-\frac{1}{32}(\pi-2)(\pi-14)$$

3. $$\sum_{n=1}^{\infty}\frac{2^nH_{n-1}}{n^2{2n \choose n}}=\frac{21}{8}\zeta(3)-\pi G$$

4. $$\sum_{n=1}^{\infty}\frac{2^n}{n(n+1)(n+2){2n \choose n}}=-\frac{1}{16}(\pi-2)(\pi-6)$$

5. $$\sum_{n=1}^{\infty}\frac{4^n}{n(n+1)(n+2){2n \choose n}}=\frac{\pi^2}{16}$$

6. $$\sum_{n=1}^{\infty}\frac{4^n}{n^2(n+1){2n \choose n}}=\frac{1}{4}(\pi-2)(\pi+2)$$

7. $$\sum_{n=1}^{\infty}\frac{4^n}{n(n+1)^2{2n \choose n}}=\frac{8-7\zeta(3)+\pi^2[\ln{4}-1]}{4}$$

8. $$\sum_{n=1}^{\infty}\frac{4^n}{(n+1)(n+2){2n \choose n}}=\frac{1}{8}(\pi^2+8)$$

9. $$\sum_{n=1}^{\infty}\frac{4^n}{(2n+1)(2n+2){2n \choose n}}=\frac{1}{8}(\pi^2-4)$$

10. $$\sum_{n=1}^{\infty}\frac{4^n}{(2n+1)(2n+2)(2n+3){2n \choose n}}=\frac{\pi^2}{8}-\frac{7}{6}$$

11. $$\sum_{n=1}^{\infty}\frac{4^n}{(2n+1)^2(2n+2){2n \choose n}}=2G-\frac{1}{2}\left(\frac{\pi^2}{4}+1\right)$$

12. $$\sum_{n=1}^{\infty}\frac{4^n}{(2n+1)(2n+2)^2{2n \choose n}}=\frac{-7\zeta(3)-4+\pi^2\ln{4}}{16}$$

13. $$\sum_{n=1}^{\infty}\frac{2^n}{(2n+1)(2n+2)(2n+3){2n \choose n}}=\frac{\pi(\pi+8)}{16}-\frac{13}{6}$$

14. $$\sum_{n=1}^{\infty}\frac{2^n n^2}{(n+2){2n \choose n}}=-\frac{1}{8}(\pi-2)(3\pi-14)$$

15. $$\sum_{n=1}^{\infty}\frac{2^n n^2}{(n+1){2n \choose n}}=-1+\pi-\frac{\pi^2}{8}$$

1. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{n+1}=\phi(2\phi-3)$$

2. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{(n+1)(n+2)}=\frac{4\phi^2-7\phi+1}{3/2}=\frac{2}{3\phi^4}$$

3. $$\sum_{n=1}^{\infty}\frac{(-16)^n}\frac{n}{(2n-1)^2}=-\frac{\ln{\phi}}{4}$$

1. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{2n-1}=\frac{1}{2}$$

2. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{(2n-1)(2n-3)}=-\frac{1}{2\phi^2}$$

3. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{(2n-1)(2n-3)(2n-5)}=\frac{1}{6\phi^4}$$

4. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{(2n-1)(2n-3)(2n-5)(2n-7)}=-\frac{1}{30\phi^6}$$

5. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{n}{(2n-1)(2n-3)\cdots [2n-(2k+1)]}=\frac{(-1)^k}{2\phi^{2k}(2k-1)!!}$$, $$k\ge 0$$

1. $$\sum_{n=1}^{\infty}\frac{4^n}\frac{1}{(n+1)^2}=3-4\ln{(2)}$$

2. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{1}{(n+1)^2}=4\phi\ln{\left(\frac{e\phi}{2}\right)}-5$$

3. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{1}{n+1}=2\phi-3$$

3. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{1}{(n+1)(n+2)}=-\frac{8\phi^2-20\phi+11}{6}$$

4. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{1}{(n+1)(n+2)(n+3)}=\frac{16\phi^3-56\phi^2+62\phi-21}{30}$$

5. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{1}{(n+1)(n+2)(n+3)(n+4)}=-\frac{128\phi^4+576\phi^3-944\phi^2+664\phi-163}{840}$$

6. $$\sum_{n=1}^{\infty}\frac{(4\phi)^n}\frac{1}{2n-1}=\phi^{-2}$$

7. $$\sum_{n=1}^{\infty}\frac{(-4\phi)^n}\frac{1}{(2n-1)(2n-3)}=\frac{\phi+\phi^{2}-\phi^{3/2}}{3\phi^{3/2}}$$

8. $$\sum_{n=1}^{\infty}\frac{(-4\phi)^n}\frac{1}{(2n-1)(2n-3)(2n-5)}=-\frac{\phi+2\phi^{2}+\phi^3-\phi^{5/2}}{15\phi^{5/2}}$$

9. $$\sum_{n=1}^{\infty}\frac{(-4\phi)^n}\frac{1}{(2n-1)(2n-3)(2n-5)(2n-7)}=\frac{\phi+3\phi^{2}+3\phi^3+\phi^4-\phi^{7/2}}{105\phi^{7/2}}$$

10. $$\sum_{n=1}^{\infty}\frac{(-4\phi)^n}\frac{1}{(2n-1)(2n-3)\cdots [2n-(2k+1)]}=(-1)^k\cdot \frac{\sqrt{\phi^{2k+1}}-\sum_{j=0}^{k}{k \choose j}\phi^{j+1}}{(2k+1)!!\sqrt{\phi^{2k+1}}}$$, $$k\ge 0$$

1. $$\sum_{n=1}^{\infty}\frac{H_n^2-H_{n,2}}{(n+1)(n+2)\cdots (n+k)}=\frac{2}{(k-1)^3(k-1)!}$$

2. $$\sum_{n=1}^{\infty}\frac{H_n^3-3H_nH_{n,2}+2H_{n,3}}{(n+1)(n+2)\cdots (n+k)}=\frac{6}{(k-1)^4(k-1)!}$$

3. $$\sum_{n=1}^{\infty}\frac{H_n^4-6H_n^2H_{n,2}+8H_{n}H_{n,3}+3H^2_{n,2}-6H_{n,4}}{(n+1)(n+2)\cdots (n+k)}=\frac{24}{(k-1)^5(k-1)!}$$

4. $$\sum_{n=1}^{\infty}\frac{H_n^2-H_{n,2}}{(n+1)^2}=2\zeta(4)$$

5. $$\sum_{n=1}^{\infty}\frac{H_n^3-3H_nH_{n,2}+2H_{n,3}}{(n+1)^2}=6\zeta(5)$$

6. $$\sum_{n=1}^{\infty}\frac{H_n^4-6H_n^2H_{n,2}+8H_{n}H_{n,3}+3H^2_{n,2}-6H_{n,4}}{(n+1)^2}=24\zeta(6)$$

7. $$\sum_{n=1}^{\infty}\frac{H_n^2-H_{n,2}}{(n+2)^2}=2[\zeta(2)+\zeta(3)+\zeta(4)]-6$$

8. $$\sum_{n=1}^{\infty}[H_n^2-H_{n,2}]\cdot\frac{2n+3}{(n+1)^2(n+2)^2}=3!-2[\zeta(2)+\zeta(3)]$$

1. $$\sum_{n=0}^{\infty}(-1)^n\left[\frac{2n \choose n}{4^n}\right]^3\cdot \frac{4n+1}{(n+1)(n+3)(2n-1)(2n-5)}=-\frac{56}{225\pi}$$

2. $$\sum_{n=0}^{\infty}\frac{(-16)^n}H_{2n}=\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^6}{25}\right)$$

3. $$\sum_{n=0}^{\infty}\frac{(-16)^n}H_{n+1}=-\phi^{-6}+\frac{1}{\sqrt{5}}\ln\left(\frac{\phi^{12}}{20^2}\right)$$

4. $$\sum_{n=0}^{\infty}\frac{(-16)^n}\left[H_{2n}-H_n\right]=\frac{1}{\sqrt{5}}\ln\left(\frac{16}{\phi^6}\right)$$

5. $$\sum_{n=0}^{\infty}\frac{(-16)^n}[H_{n+1}-H_{n}]=-\phi^{-6}$$

1. $$-\frac{4\Gamma^2\left(\frac{3}{4}\right)+\Gamma^2\left(\frac{1}{4}\right)}{2\pi\sqrt{2\pi}}=\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2\cdot\frac{(-1)^n}{2n-1}$$

2. $$\frac{4\Gamma^2\left(\frac{3}{4}\right)}{\pi\sqrt{2\pi}}=\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2\cdot\frac{(-1)^n}{(2n-1)^2}$$

3. $$-\sqrt{10(\pi-3)}\frac{\Gamma^2\left(\frac{1}{4}\right)-4\Gamma^2\left(\frac{3}{4}\right)}{8\pi\sqrt{\pi}}\approx\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2\cdot\frac{(-1)^n n}{2n-1}$$

1. $$\sum_{n=1}^{\infty}\left[\frac{2n \choose n}{4^n}\right]^2\cdot \frac{1}{(2n-1)^2}=\frac{4}{\pi}-1$$

2. $$\sum_{n=1}^{\infty}\left[\frac{2n \choose n}{4^n}\right]^2\cdot \frac{n}{(2n-1)^2}=\frac{1}{\pi}$$

3. $$\sum_{n=1}^{\infty}\left[\frac{2n \choose n}{4^n}\right]^2\cdot \frac{1}{(2n-1)}=1-\frac{2}{\pi}$$

4. $$\sum_{n=1}^{\infty}\left[\frac{2n \choose n}{4^n}\right]^2\cdot \frac{1}{n(2n-1)^2}=\frac{4-\pi}{\pi}$$

5. $$\sum_{n=1}^{\infty}\left[\frac{2n \choose n}{4^n}\right]^2\cdot \frac{n}{(2n-1)^2(n+1)(n+2)\cdots (n+k)}= \frac{4^kk!}{(2k+1)!!^2}\cdot \frac{1}{\pi}$$, $$k\ge 1$$

6. $$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{32^n}\cdot \frac{n}{(2n-1)^2(n+1)(n+2)\cdots (n+k)}=\frac{2^{k-2}}{\sqrt{2\pi}(2k+)!!}\cdot \frac{\Gamma\left(\frac{2k+3}{4}\right)}{\Gamma\left(\frac{2k+5}{4}\right)}$$, $$k\ge 1$$

7. $$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{32^n}\cdot \frac{n}{(2n-1)^2}=\frac{1}{4\sqrt{2\pi}}\cdot \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}$$

8. $$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{32^n}\cdot \frac{n^2}{(2n-1)^2}=\frac{1}{8\sqrt{2\pi}}\cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$

9. $$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{32^n}\cdot \frac{n}{2n-1}=\frac{\Gamma^2\left(\frac{1}{4}\right)-4\Gamma^2\left(\frac{3}{4}\right)}{8\pi\sqrt{\pi}}$$

10. $$\sum_{n=1}^{\infty}\frac{{2n \choose n}^2}{32^n}\cdot \frac{1}{2n-1}=\frac{\Gamma^2\left(\frac{1}{4}\right)+4\Gamma^2\left(\frac{3}{4}\right)}{4\pi\sqrt{\pi}}$$

1. $$\sum_{n=1}^{\infty}\frac{2^n}\cdot \frac{1}{n^2(n+1)}=\frac{\pi-2}{2}$$

2. $$\sum_{n=1}^{\infty}\frac{2^n}\cdot \frac{1}{n(n+1)}=\frac{\pi^2-4\pi+8}{8}$$

3. $$\sum_{n=1}^{\infty}\frac{2^n}\cdot \frac{1}{n^2(n+2)}=-\frac{(\pi-14)(\pi-2)}{32}$$

4. $$\sum_{n=1}^{\infty}\frac{2^n}\cdot \frac{1}{n^2(n+2)}=-\frac{(5\pi-12)(275\pi-1292)}{8000}$$

5. $$\sum_{n=1}^{\infty}\frac{2^n}\cdot \frac{1}{n^2(n+1)(n+2)}=\frac{\pi^2-4}{32}$$

6. $$\sum_{n=1}^{\infty}\frac{4^n}\cdot \frac{1}{n^2(n+1)}=\frac{\pi^2-4}{4}$$

7. $$\sum_{n=1}^{\infty}\frac{4^n}\cdot \frac{1}{n(n+1)}=\frac{\pi^2+4}{4}$$

8. $$\sum_{n=1}^{\infty}\frac{4^n}\cdot \frac{1}{n(n+1)(n+2)}=\frac{\pi^2}{16}$$

1. $$\sum_{n=0}^{\infty}\frac{2^nH_n}{(2n+1){2n \choose n}}=2G-\frac{1}{2}\pi\ln{2}$$

2. $$\sum_{n=0}^{\infty}\frac{2^nH_n}{n(2n+1){2n \choose n}}=-2G+\frac{\pi(\pi+4\ln{2})}{8}$$

3. $$\sum_{n=0}^{\infty}\frac{2^nH_n}{n{2n \choose n}}=2G+\frac{\pi(\pi-4\ln{2})}{8}$$

4. $$\sum_{n=0}^{\infty}\frac{2^nH_n}=2G+\pi\left(1-\frac{1}{2}\ln{2}\right)$$

5. $$\sum_{n=2}^{\infty}\frac{4^n}\cdot \frac{1}{n(n-1)^2}=\pi^2-8$$

6. $$\sum_{n=2}^{\infty}\frac{4^n}\cdot \frac{1}{n^2(n-1)^2}=\frac{3\pi^2-28}{2}$$

7. $$\sum_{n=2}^{\infty}\frac{4^n}\cdot \frac{1}{n^2(n-1)}=\frac{12-\pi^2}{2}$$

8. $$\sum_{n=2}^{\infty}\frac{2^n}\cdot \frac{1}{n(n-1)}=\frac{4-\pi}{2}$$

9. $$\sum_{n=2}^{\infty}\frac{2^n}\cdot \frac{1}{(n-1)}=1$$

10. $$\sum_{n=2}^{\infty}\frac{2^n}\cdot \frac{1}{(n-1)^2}=\frac{\pi^2+4\pi-16}{8}$$

11. $$\sum_{n=2}^{\infty}\frac{2^n}\cdot \frac{1}{n^2(n-1)}=\frac{24-\pi^2-4\pi}{8}$$

12. $$\sum_{n=2}^{\infty}\frac{2^n}\cdot \frac{1}{n^2(n-1)^2}=\frac{\pi^2+6\pi-28}{4}$$

1. $$\sum_{n=0}^{\infty}\frac{4^n}\dot \frac{1}{(2n+1)^2-(2k)^2}=-\frac{(2k-1)!!}{(2k)!!}\cdot \frac{\pi}{8k}$$, $$k\ge 1$$

2. $$\sum_{n=0}^{\infty}\frac{H_n}{(2n+1)^2-(2)^2}=\frac{1}{2}$$

3. $$\sum_{n=0}^{\infty}\frac{H_n^2}{(2n+1)^2-(2)^2}=\frac{2}{2}$$

3. $$\sum_{n=0}^{\infty}\frac{H_n^3}{(2n+1)^2-(2)^2}=\frac{4+\zeta(2)}{2}$$

4. $$\sum_{n=0}^{\infty}\frac{H_{n}^{(2)}}{(2n+1)^2-(2)^2}=\ln{4}-1$$

5. $$\sum_{n=0}^{\infty}\frac{H_n}{(2n+1)^3-(2n+1)}=\left[\frac{\ln{2}}{2}\right]^2$$

6. $$\sum_{n=0}^{\infty}\frac{H_n}{(2n)^2-1}=\ln{2}$$

7. $$\sum_{n=0}^{\infty}\frac{nH_n}{((2n)^2-1)^2}=\frac{\pi^2}{32}-\frac{1}{2}\ln{2}$$

8. $$\sum_{n=0}^{\infty}\frac{H_n}{n[(2n)^2-1]}=2\ln{2}(1-\ln{2})$$

9. $$\sum_{n=0}^{\infty}\frac{[4n^2+4n-1]H_n}{[(2n)^2-1]^2}=\frac{\pi^2}{8}$$

1. $$\sum_{n=0}^{\infty}\frac{32^{-n}{2n \choose n}^2}{(2n-1)(n+1)^2(n+2)...(n+k)}=F(k)$$

2. $$\sum_{n=0}^{\infty}\left[\frac{4^n}\right]^2\cdot \frac{H_n}{2n+1}=\frac{4}{\pi}-1$$

3. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{[2n-(2k-1)]^4-[2n+(2k+1)]^4}=-\frac{\pi[k\pi-\tanh{(k\pi)]}}{2^9\times k^4}$$, $$k\ge 1$$

4. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(2n+1)^k}=\frac{2^{k+1}-1}{2^{k+1}}\zeta(k+1)$$, $$k\ge 1$$

5. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(n+1)}=\ln{4}$$

6. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(2n+3)}=\frac{1}{2}$$

7. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(n+k)}=\frac{H_{k}-\ln{4}}{2k-1}$$, $$k\ge 1$$

8. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(2n+1)[(2n+1)^2+(2k)^2]}=\frac{\pi[k\pi-\tanh{(k\pi)]}}{2^5\times k^3}$$, $$k\ge 1$$

9. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(2n+1)[(2n+1)^2-(2k)^2]}=-\frac{\pi^2}{2^5k^2}$$, $$k\ge 1$$

10. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{1}{(2n+1)[(2n+1)^4-(2k)^4]}=\frac{\pi[k\pi-\tanh{(k\pi)]}}{2^7\times k^5}$$, $$k\ge 1$$

11. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{2n+1}{[(2n+1)^4-(2k)^4]}=\frac{-\pi\tanh{(k\pi)}}{2^6\times k^3}$$, $$k\ge 1$$

12. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{4^n}\frac{2n+1}{[(2n+1)^2-(2k)^2]}=0$$, $$k\ge 1$$

1. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{4^n}\cdot \frac{1}{(2n-1)^2(2n+3)^2}=\frac{1}{8}$$

2. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{4^n}\cdot \frac{1}{(2n-3)^2(2n+5)^2}=\frac{1}{72}$$

3. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{4^n}\cdot \frac{1}{(2n-1)^3(2n+3)^3}=\frac{7\zeta(3)-26}{512}$$

4. $$\sum_{n=0}^{\infty}\frac{2^nH_{n-1}^{(2)}}{n^2{2n \choose n}}=\frac{\pi^4}{384}$$

5. $$\sum_{n=0}^{\infty}\left(\frac{1}{x^2}\right)^n\frac{H_{n-1}^{(2)}}{n^2{2n \choose n}}=\frac{2}{3}\arcsin^4\left(\frac{1}{2x}\right)$$

6. $$\sum_{n=0}^{\infty}\left(-\frac{1}{x^2}\right)^n\frac{H_{n-1}^{(2)}}{n^2{2n \choose n}}=\frac{2}{3}arcsinh^4\left(\frac{1}{2x}\right)$$

7. $$\sum_{n=0}^{\infty}\frac{32^{-n}{2n \choose n}^2}{(2n-1)(n+1)(n+2)\cdots (n+k)}=-\frac{2^{k-1}}{(2k-1)!!\sqrt{2\pi}}\left[\frac{\Gamma\left(\frac{2k+1}{4}\right)}{\Gamma\left(\frac{2k+3}{4}\right)}+\frac{\Gamma\left(\frac{2k+3}{4}\right)}{\Gamma\left(\frac{2k+5}{4}\right)}\right]$$

8. $$\sum_{n=0}^{\infty}\frac{4^{-n}{2n \choose n}}{(2n-1)(n+1)(n+2)\cdots (n+k)}=F(k)$$

1. $$\sum_{n=0}^{\infty}C_i[(2n-1)\pi]=\frac{5}{8}(2\gamma-1)$$

2. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{2n-1}\cdot 4^{-n}=-\frac{1}{2}$$

3. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{2n-3}\cdot 4^{-n}=-\frac{1}{3}$$

4. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{4^n}\cdot \frac{1}{(2n-1)(2n-3)\cdots (2n-(2k+1))}=-(-1)^k\frac{1}{(2k+2)(2k+1)!!}$$

5. $$\sum_{n=0}^{\infty}\frac{2^nnH_n}{(2n+1){2n \choose n}}=\frac{\pi}{2}$$

6. $$\sum_{n=0}^{\infty}\frac{2^nnH_n^{(2)}}{(2n+1){2n \choose n}}=\frac{\pi^2}{8}$$

7. $$\sum_{n=0}^{\infty}\frac{2^nH_n}=2C+\pi-\frac{1}{2}\pi \ln{2}$$

8. $$\sum_{n=0}^{\infty}\frac{2^nH_n}{{2n \choose n}(2n+1)}=2C-\frac{1}{2}\pi \ln{2}$$

9. $$\sum_{n=0}^{\infty}\frac{2^nH_n}{(2n+1)C_n}=2C+\frac{1}{2}\pi(1-\ln{2})$$

10. $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot \frac{4^n}(2n-1)^{-2}=\frac{12+\pi^2}{16}$$

1. $$\sum_{n=0}^{\infty}\left[\frac{\sin(2\theta)}{2}\right]^{2n}\cdot\frac{(2n)^3-2n}=\frac{1}{2}\cos(2\theta)-xcosec(2\theta)-\ln{\cos(\theta)}$$

2. $$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{n^4}{[(2n)^3-2n]^2}=\frac{1}{128}\pi\ln{2}$$

3. $$\sum_{n=0}^{\infty}\frac{4^n}\cdot \frac{n^2}{(2n)^3-2n}=\frac{\pi}{16}$$ 4. $$\sum_{n=0}^{\infty}\frac{1}{x^n}\cdot\frac{(2n-1)!!}=e^{2/x}$$

5. $$\sum_{n=0}^{\infty}\frac{1}{x^n}\cdot\frac{(2n-1)!!^2}=cosh\left(\frac{2}{\sqrt{x}}\right)$$

6. $$\sum_{n=0}^{\infty}\frac{1}{(-x)^n}\cdot\frac{(2n-1)!!^2}=\frac{1}{\sqrt{x}}sinh\left(\frac{2}{\sqrt{x}}\right)$$

7. $$\sum_{n=0}^{\infty}\frac{1}{x^n}\cdot\frac{(2n-1)!!^2}=cosh\left(\frac{4}{\sqrt{x}}\right)I_0\left(\frac{4}{\sqrt{x}}\right)$$

8. $$\sum_{n=0}^{\infty}\frac{1}{(-x)^n}\cdot\frac{(2n-1)!!^2}=\cos\left(\frac{4}{\sqrt{x}}\right)J_0\left(\frac{4}{\sqrt{x}}\right)$$

9. $$\sum_{n=0}^{\infty}\frac{H_{n,k}}{(-x^2)^n}\cdot\frac{1}{(n)!^2}=\zeta(k)J_0\left(\frac{2}{x}\right)$$

10. $$\sum_{n=0}^{\infty}\left(\pm\frac{1}{x}\right)^n\frac{(2n-1)!!}=e^{\pm4/x}I_0(4/x)$$

11. $$\sum_{n=0}^{\infty}\left(\frac{1}{x^2}\right)^n\frac{{2n \choose n}^2}{((2n-1)!!)^2}=I_0(4/x)$$

12. $$\sum_{n=0}^{\infty}\left(-\frac{1}{x^2}\right)^n\frac{{2n \choose n}^2}{((2n-1)!!)^2}=J_0(4/x)$$

1. $$\sum_{n=0}^{\infty}(-1)^n\frac{n}{2n-1}\left[\frac{4^n}\right]^2=\frac{\sqrt{2\pi}}{\Gamma^2(1/4)}$$

2. $$\sum_{n=0}^{\infty}(-1)^n\frac{n}{(2n-1)(n+1)^2}\left[\frac{4^n}\right]^2=\frac{1}{9}\left[12-\frac{1}{\pi}\sqrt{\frac{2}{\pi}}\left(\Gamma^2\left(\frac{1}{4}\right)+24\Gamma^2\left(\frac{3}{4}\right)\right)\right]$$

3. $$\sum_{n=0}^{\infty}\frac{n^2}{(2n-1)^3}\left[\frac{4^n}\right]^2=\frac{C}{\pi}$$

4. $$\sum_{n=0}^{\infty}\frac{n}{(4n^2-1)}\left[\frac{4^n}\right]^2=\frac{2C-1}{\pi}$$

5. $$\sum_{n=0}^{\infty}\frac{n}{(2n-1)^2(n+1)}\left[\frac{4^n}\right]^2=\frac{4}{9\pi}$$

6. $$\sum_{n=0}^{\infty}\frac{1}{(4n^2-1)}\left[\frac{4^n}\right]^2=\frac{1-2C}{\pi}$$

7. $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n-1)(n+1)^2(n+2)}\left[\frac{4^n}\right]^2=-\frac{4}{3}+\frac{2^6\sqrt{2\pi}}{15\Gamma^2(1/4)}$$

8. $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n-1)(n+1)(n+2)^2}\left[\frac{4^n}\right]^2=\frac{2^4}{15^2}\left[\frac{1}{3\pi}\sqrt{\frac{2}{\pi}}\left(54\Gamma^2\left(\frac{3}{4}\right)-5\Gamma^2\left(\frac{1}{4}\right)\right)-5\right]$$

9. $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n-1)(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=\frac{2^2}{15^2}\left[\frac{4}{3\pi}\sqrt{\frac{2}{\pi}}\left(36\Gamma^2\left(\frac{3}{4}\right)+5\Gamma^2\left(\frac{1}{4}\right)\right)-55\right]$$

10. $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n-1)^2(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=\frac{2^2}{3\times 5^3}\left[35-\frac{2^6\pi}{\Gamma^2\left(\frac{1}{4}\right)}\sqrt{\frac{2}{\pi}}\right]$$

11. $$\sum_{n=0}^{\infty}(-1)^n\frac{16n^3+64n^2+49n-15}{(2n-1)(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=\frac{16}{15}\left[25-\frac{116\sqrt{2\pi}}{\Gamma^2\left(\frac{1}{4}\right)}\right]$$ (2)+(8)

12. $$\sum_{n=0}^{\infty}(-1)^n\frac{16n^3+64n^2+64n+15}{(2n-1)(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=\frac{4}{15}\left[25-\frac{224\sqrt{2\pi}}{\Gamma^2\left(\frac{1}{4}\right)}\right]$$ (2)+(9)

13. $$\sum_{n=0}^{\infty}(-1)^n\frac{16n^3+64n^2+64n+25}{(2n-1)(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=-\frac{192}{5}\frac{\sqrt{2\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$ (11)-(12)

14. $$\sum_{n=0}^{\infty}(-1)^n\frac{8n^2+12n+7}{(2n-1)^2(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=4$$ (7)+(10)

15. $$\sum_{n=0}^{\infty}(-1)^n\frac{14n^2+21n+11}{(2n-1)^2(n+1)^2(n+2)^2}\left[\frac{4^n}\right]^2=\frac{64}{5}\cdot \frac{\sqrt{2\pi}}{\Gamma^2\left(\frac{1}{4}\right)}$$ (7)+(10)

16. $$\sum_{n=0}^{\infty}\frac{H_n}{x^n(n+1)}=\frac{x}{2}\ln^2\left(\frac{x}{x-1}\right)$$

17. $$\sum_{n=0}^{\infty}\frac{(-1)^nH_n}{x^n(n+1)}=-\frac{x}{2}\ln^2\left(\frac{x+1}{x}\right)$$

18. $$\sum_{n=0}^{\infty}\frac{H_n{2n \choose n}}{4^n(2n-1)(2n-3)\cdots [2n-(2k+1)]}=\frac{2(-1)^k}{(2k+1)!!(2k+1)}$$