User:Series and sum 123

14/2/21-21/4/21:

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n+1}}\color{green}{\left(\frac{n+1}{2n-1}\right)^4}(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=-\color{red}{\frac{1}{\pi^2}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n+1}}\color{brown}{\left(\frac{n+1}{2n-1}\right)^4}(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k-1)k}{(2k-1)(2k+1)}=\color{red}{\frac{1}{\pi^2}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n+2}}\color{blue}{\left(\frac{n+1}{2n-1}\right)^4}(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^2-2k+1)}{(2k-1)(2k+1)}=\color{red}{\frac{1}{\pi^2}}$$

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

$$ n=1,2,3...,1^2,2^2,8^2,40^2,224^2,1344^2,...=2^{2n}C_{n}^2$$

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n+1}}\color{green}{\left(\frac{n+1}{2n-1}\right)^4}(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(k-n)[4(j-1)k^2+k-j]}{(2k-1)(2k+1)}=\color{red}{\frac{1}{\pi^2}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n+1}}\color{green}{\left(\frac{n+1}{2n-1}\right)^4}(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(k-n)(4\phi k^2+k-\phi^2)}{(2k-1)(2k+1)}=\color{red}{\frac{1}{\pi^2}}$$

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$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2k^2}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{576\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2(1-6k)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{3\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2(51k+8)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{32\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2(22k+39)k}{(2k-1)(2k+1)(2k-3)}=-\color{red}{\frac{1}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2(22k^2+3k+6)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)^2(3282k+509)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{\pi}}$$

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$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(n+k^2+k-4)k}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{1350\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^3+16k^2+(n+236)k-60)}{(2k-1)(2k+1)(2k-3)}=\color{blue}{\frac{1}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^3-884k^2+(33n-9734)k+3540)}{(2k-1)(2k+1)(2k-3)}=\color{blue}{\frac{1}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(-7k+27)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(135k^2-527k+135)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(10k^2+170k-57)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(540k^2-2101k+513)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(180k^2-869k+243)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(40k^2-153k+60)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(90k^3+25k^2+227k-84)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{\pi}}$$

$$F(p,k)=k^3(20-70p)+25k^2(3p+4)+k(373p+600)-12(11p+18)$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)F(p,k)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{\pi}}$$

$$F(p,k)=k^3(20-70p)+25k^2(3p+4)+k(373p+600)-12(11p+18)$$

$$F(p,k)=k^3(28-62p)+k^2(23p+48)+k(88-139p)-6(5-9p)$$

$$F(p,k)=k^3(4+4p)+k^2(-26p-26)+k(-256-256p)-3(-31-31p)$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(14k^3+24k^2+44k-15)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{2\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(-34k^3+71k^2-51k+24)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^3-26k^2-256k+93)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{0}{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\color{green}{\left(\frac{(n+1)^{3/2}}{(2n-1)(2n-3)(2n-5)}\right)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^3-163k^2-2355k+861)}{(2k-1)(2k+1)(2k-3)}=\color{red}{\frac{1}{\pi}}$$

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$$ P_{n+1}P_{n+k}-P_1P_2...P_n=P_x,k\ge1$$

$$P_x$$ are most likely be prime

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$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)((P_{z+3}+3)k^2-F(P_{z+3})k+(P_{z+3}+3))}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi};z\ge1$$

$$P_z=$$ Prime numbers

$$ 7,11,19,23,31,43,47,59,67,... $$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(10k^2-83k+10)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(14k^2-145k+14)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(22k^2-269k+22)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(26k^2-331k+26)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(34k^2-455k+34)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(46k^2-641k+46)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(50k^2-703k+50)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(62k^2-889k+62)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(70k^2-1013k+70)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

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$$A_n(a,b,c)=\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(ak^2+bk+c)}{(2k-1)(2k+1)(2k-3)}$$

$$ \sum_{n=0}^{\infty}(4n^2-n+20)A_n(6,-97,6)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^2-11k+1)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{16\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-23k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-19k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{6\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-13k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{4\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-7k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+5k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(10k^2-119k+10)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2+41k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(10k^2-83k+10)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(14k^2-145k+14)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(22k^2-269k+22)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(26k^2-331k+26)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(34k^2-455k+34)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-25k+2)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{12\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(14k^2-181k+14)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(10k^2-131k+10)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2(n+1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^2-53k+4)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{8\pi}$$

$$A_n=\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2\frac{n+1}{2n-7}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)k}{(2k-1)(2k+1)(2k-3)}$$

Where $$p=$$ prime numbers.

If $$p\ge11$$,

then the expression $$(24np-49p+210)$$ will most likely not have a factor.

$$ \sum_{n=0}^{\infty}(24np-49p+210)A_n=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}(24np-49p-210)A_n=\frac{1}{\pi}$$

Conjecture: Where $$p_a$$ is prime and $$p_b$$ are mostly like will be prime. If

$$7p_a-30=p_b$$

$$A_n(a,b,c)=\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{(2n-1)(2n-3)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(ak^3+bk^2+ck+d)}{(2k-1)(2k+1)(2k-3)}$$

$$\sum_{n=0}^{\infty}(8n^2+n-12)A_n(0,0,0,1)=\frac{1}{36\pi}$$

$$ \sum_{n=0}^{\infty}(4n^2-n+5)A_n(0,0,1,0)=\frac{1}{10\pi}$$

$$ \sum_{n=0}^{\infty}(4n^2-n+20)A_n(0,0,1,0)=\frac{1}{6\pi}$$

$$ \sum_{n=0}^{\infty}(n+1)A_n(0,0,1,0)=\frac{1}{72\pi}$$

$$ \sum_{n=0}^{\infty}n(n+1)A_n(0,0,1,0)=\frac{1}{32\pi}$$

$$ \sum_{n=0}^{\infty}(8n^2+n-12)A_n(0,60,20,767)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(8n^2+n-12)A_n(8,-8,8,47)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(8n^2+n-12)A_n(4,-6,6,-11)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}(8n^2+n-12)A_n(0,0,10,-11)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}(8n^2+n-12)A_n(0,0,20,58)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(-4n^2+n+18)A_n(0,0,0,1)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(4n^2-n-18)A_n(0,2,-255,-2)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(-4n^2+n+18)A_n(0,2,-255,0)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(32n^2-8n+85)A_n(0,0,1,0)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}(112n-13)A_n(0,0,1,0)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(n+1)A_n(0,0,1,0)}{2n-7}=-\frac{1}{210\pi}$$

$$\sum_{n=0}^{\infty}\frac{(8n-1)(n-4)A_n(0,0,1,0)}{2n-7}=\frac{1}{21\pi}$$

$$ \sum_{n=0}^{\infty}\frac{(8n^2-233n-196)A_n(0,0,1,0)}{2n-7}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(n+1)(24n+161)A_n(0,0,1,0)}{2n-7}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(n+1)(264n-329)A_n(0,0,1,0)}{2n-7}=-\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}\frac{(n+1)(24np-49p+210)A_n(0,0,1,0)}{2n-7}=-\frac{1}{\pi}$$

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$$A_n=\frac{C_n^3}{2^{6n}}\left(\frac{\sqrt{n^2-1}}{(2n-1)(2n-3)(2n-5)}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{n-k}{(2k-1)(2k+1)(2k-3)}$$

$$\sum_{n=0}^{\infty}A_n(186n^2-761n+1111)=\frac{1}{\pi}$$

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$$ A_n=\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{n-k}{(2k-1)(2k+1)(2k-3)}$$

$$\sum_{n=0}^{\infty}\frac{(n-1)(n-2)}{(2n-3)^2}A_n=\frac{1}{432\pi}$$

$$\sum_{n=0}^{\infty}\frac{8n^2-22n+27}{(2n-3)^2}A_n=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{-6n^2+131n-125}{(2n-3)^2(2n-5)^2}A_n=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(n-1)(361n-1085)}{(2n-3)^2(2n-5)^2(2n-7)}A_n=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{7(n-1)(n+40)}{(2n-3)^2(2n-5)^2(2n-7)}A_n=\frac{1}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(6k^3-5k^2+2k-6)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(-2k^3+3k^2+62k+1)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(9n^2-20n+9)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=-\frac{1}{16\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(n-1)(n-2)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{432\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(36n^2-161n+144)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(36n^2-65n+36)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(36n^2-89n+36)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(108n^2-219n+108)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{(n+1)}{(2n-1)(2n-3)}\right)^2(12n^2-35n+36)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(ak^2+bk+c)}{(2k-1)(2k+1)(2k-3)}=F(a,b,c)$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(ak^3+bk^2+cx+d)}{(2k-1)(2k+1)(2k-3)}=G(a,b,c,d)$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k^2+k-5)}{(2k-1)(2k+1)(2k-3)}=\frac{1}{144\pi}$$

$$ \sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(2k^2-2k-25)k}{(2k-1)(2k+1)(2k-3)}=\frac{1}{144\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\left(\frac{n+1}{2n-1}\right)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)k}{(2k-1)(2k+1)(2k-3)}=\frac{1}{72\pi}$$

$$ F(12,12,-49)=\frac{1}{\pi}$$

$$ F(38,79,-188)=\frac{1}{\pi}$$

$$ G(2,27,31,-140)=\frac{1}{\pi}$$

$$G(2,-1,47,-5)=\frac{1}{\pi}$$

$$ G(2,-1,17,0)=\frac{1}{\pi}$$

$$ G(6,-3,-123,5)=\frac{1}{\pi}$$

$$ G(4,0,-127,5)=\frac{1}{\pi}$$

$$ G(8,-2,-117,-15)=\frac{1}{\pi}$$

$$F(44,42,-211)=\frac{1}{\pi}$$

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$$A_n=\frac{C_n^3}{2^{6n}}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}$$

$$ B_n=\frac{C_n^3}{2^{6n}}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)(2k-3)}$$

$$ C_n=\frac{C_n^3}{2^{6n}}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(-4k^2+2k-1)}{(2k-1)(2k+1)(2k-3)}$$

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

$$\sum_{n=0}^{\infty}\frac{(n+1)^2}{(2n-1)^2}B_n=-\frac{5}{72\pi}$$

$$\sum_{n=0}^{\infty}\frac{(n+1)^2}{(2n-1)^2}C_n=-\frac{61}{36\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_n^3}{2^{6n}}\frac{(n+1)^2}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)(4k^2+2k+1)}{(2k-1)(2k+1)(2k-3)}=\frac{7}{4\pi}$$

-

$$ z_{n-j}=\frac{C_n}{2^{2n+2}}\cdot \frac{\left(\frac{1}{2}\right)_{n-j}}{(1)_{n-j}}$$

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

$$\sum_{n=0}^{\infty}z_{n+j}=\frac{(2j-3)!!}{2(2j-2)!!}-\frac{1}{(2j-1)\pi}$$

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$$ z_n=\frac{C_n}{2^{2n+2}}\cdot \frac{\left(\frac{1}{2}\right)_n}{(1)_n}$$

$$\sum_{n=0}^{\infty}z_n(n+1)\prod_{j=0}^{v}\frac{1}{2n-2j-1}=\frac{(2v)!!}{(2v+1)!!^2}\cdot \frac{(-1)^{v+1}}{2\pi}$$

$$\sum_{n=0}^{\infty}z_n(n+1)\prod_{j=0}^{v}\frac{1}{(2n-2j-1)^2}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^3\cdot \frac{v!^2}{\pi}$$

$$\sum_{n=0}^{\infty}z_n(n+1)\sum_{j=0}^{v}(-1)^{j+1}{2v \choose j}\frac{1}{2n-2j-1}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^2\cdot \frac{1}{2\pi}$$

$$\sum_{n=0}^{\infty}z_n(n+1)\left(\sum_{j=0}^{v}(-1)^{j}{2v \choose j}\frac{1}{2n-2j-1}\right)^2=\left(\frac{(2v)!!}{(2v+1)!!}\right)^3\cdot \frac{2^{2v}}{\pi}$$

$$\sum_{n=0}^{\infty}z_n\prod_{j=0}^{v}\frac{1}{2n-2j-1}=\frac{(2v)!!}{(2v+1)!!^2}\cdot \frac{2v+2}{2v+3}\cdot\frac{(-1)^{v}}{\pi}$$

$$\sum_{n=0}^{\infty}z_n\sum_{j=0}^{v}(-1)^{j+1}{2v \choose j}\frac{1}{2n-2j-1}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^2\cdot\frac{2v+2}{2v+3}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}z_n\left(\sum_{j=0}^{v}(-1)^{j}{2v \choose j}\frac{1}{2n-2j-1}\right)^2=\left(\frac{(2v)!!}{(2v+1)!!}\right)^3\cdot\frac{2^{2v+3}}{(2v+3)^2}\cdot\frac{v+1}{\pi}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_{n}^3}{(1)_{n}^3}$$

$$ z_n=\frac{C_n}{16^n}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(n-1)}{(2k-1)(2k+1)}$$

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

$$ C_n=\frac{C_n}{(-32)^{2n}}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(n-1)}{(2k-1)(2k+1)}$$

$$\sum_{n=0}^{\infty}4A_nu_{n}=\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}$$

$$\sum_{n=0}^{\infty}4z_nu_{n}=\frac{3\sqrt{3}\Gamma\left(\frac{1}{3}\right)^2\Gamma\left(\frac{7}{6}\right)^2}{\pi^3}$$

$$\sum_{n=0}^{\infty}4u_nC_n=\frac{32\sqrt{2}\Gamma(5/4)^4}{\pi^3}$$

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$$ u_n=\frac{\left(\frac{1}{2}\right)_n^5}{(1)_n^5}$$

$$ v_n=\frac{\left(\frac{1}{2}\right)_n^7}{(1)_n^7}$$

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

$$C_n$$ is Catalan numbers.

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_nA_n}{2^{4n}}\left(20n^2+8n+1\right)=\frac{2}{\pi^2}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_nA_n}{2^{12n}}\left(820n^2+180n+13\right)=\frac{32}{\pi^2}$$

$$\sum_{n=0}^{\infty}\frac{v_nA_n}{2^{8n}}\left(168n^3+76n^2+14n+1\right)=\frac{8}{\pi^2}$$

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

Example 1:

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^5}{(1)_n^5}$$

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

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{2^{2n+1}}\left(\frac{(20j^2-8j+1)A_n}{2^{2n}}+(5n-5j+2)(n+j)\right)=\frac{1}{\pi^2}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{2^{2n+1}}\left(\frac{(20j^2+8j+1)A_n}{2^{2n}}+(5n+5j+2)(n-j)\right)=\frac{1}{\pi^2}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{2^{2n+1}}\left(\frac{(20j^2+1)A_n}{2^{2n}}+5n^2+2n-5j^2\right)=\frac{1}{\pi^2}$$

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

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$$ u_n=\frac{\left(\frac{1}{2}\right)_n^5}{(1)_n^5}$$

$$ w_n=\frac{\left(\frac{1}{2}\right)_n^7}{(1)_n^7}$$

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

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{2^{2n+1}}\left(\frac{A_n}{2^{2n}}+5n^2+2n\right)=\frac{1}{\pi^2}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{2^{10n+5}}\left(\frac{13A_n}{2^{2n}}+205n^2+45n\right)=\frac{1}{\pi^2}$$

$$\sum_{n=0}^{\infty}\frac{w_n}{2^{6n+4}}\left(\frac{A_n}{2^{2n-1}}+84n^3+38n^2+7n\right)=\frac{1}{\pi^3}$$

$$\sum_{n=0}^{\infty}\sum_{j=0}^{n}{n \choose j}^4\frac{1}{6^{2n+1}}\left(\frac{A_n}{2^{2n}}+n\right)=\frac{\sqrt{15}}{20\pi}$$

$$\sum_{n=0}^{\infty}{2n \choose n}^2\sum_{j=0}^{n}{2n\choose j}^2{2n-2j \choose n-j}^2\frac{1}{2^{10n+3}}\left(\frac{A_n}{2^{2n}}+9n^2+3n\right)=\frac{1}{\pi^2}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_n\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{882^{2n+1}}\left(\frac{1123A_n}{4^{n}}+5365n\right)=\frac{1}{\pi}$$

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

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{18^{2n+1}}\left(\frac{23A_n}{4^{n}}+65n\right)=\frac{1}{\pi}$$

General of Ramanujan;

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

$$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{18^{2n+1}}\left(260n+23\right)=\frac{4}{\pi}$$

Transform into: $$\sum_{n=0}^{\infty}(-1)^n\frac{u_n}{18^{2n+1}}\left(\frac{23B_n}{z^{2n}}+\frac{260n}{z^2}\right)=\frac{4}{z^2\pi}$$

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$$ u_n=\frac{\left(\frac{1}{2}\right)_n\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}\frac{u_n}{6^{2n}}\left(\frac{A_n}{4^n}+2n\right)=\frac{\sqrt{3}}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{u_n}{7^{4n+2}}\left(\frac{3A_n}{4^{n}}+10n\right)=\frac{\sqrt{3}}{36\pi}$$

$$\sum_{n=0}^{\infty}\frac{u_n}{9^{2n+1}}\left(\frac{2A_n}{4^{n}}+5n\right)=\frac{\sqrt{3}}{8\pi}$$

$$\sum_{n=0}^{\infty}\frac{u_n}{99^{4n+2}}\left(\frac{2206A_n}{4^{n}}+13195n\right)=\frac{\sqrt{2}}{8\pi}$$

$$\sum_{n=0}^{\infty}\frac{u_n}{99^{2n+1}}\left(\frac{19A_n}{4^{n}}+70n\right)=\frac{\sqrt{11}}{22\pi}$$

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

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$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

$$A_n=\frac{C_n}{16^n}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}$$

$$B_n=\frac{C_n}{16^{2n}}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}$$

$$ C_n=\frac{C_n}{(-32)^{n}}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}$$

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

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

$$\sum_{n=0}^{\infty}u_n\left(10B_n+\frac{21n}{64^n}\right)=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}u_n\left(2C_n+\frac{3n}{(-8)^n}\right)=\frac{\sqrt{2}}{\pi}$$

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

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

$$ u_n=\frac{\left(\frac{1}{2}\right)_n}{(1)_n}$$

$$ z_n=\frac{C_n^2}{16^n}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}$$

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

$$\sum_{n=0}^{\infty}\frac{z_n}{n+2}=\frac{1}{6}$$

$$\sum_{n=0}^{\infty}\frac{z_n}{2n-1}=-\frac{1}{6}$$

$$\sum_{n=0}^{\infty}\frac{16n(n+1)}{(2n-1)^2}z_n=\pi$$

$$\sum_{n=0}^{\infty}\frac{8n(n+1)}{(2n-3)^2}z_n=\pi$$

$$\sum_{n=0}^{\infty}\frac{6n(n+1)}{(2n-5)^2}z_n=\pi$$

$$\sum_{n=0}^{\infty}\frac{5n(n+1)}{(2n-7)^2}z_n=\pi$$

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

$$\sum_{n=0}^{\infty}u_nz_n=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n\frac{3-7n}{(2n-1)(2n-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n\frac{3-28n^2}{(2n-1)(2n-3)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n\frac{n(56n+1)}{(2n-1)(2n-3)}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n\prod_{j=0}^{v}\frac{1}{2n-2j-1}=\frac{(-1)^{v+1}}{\pi}\cdot\frac{(2v)!!}{(2v+1)!!^2}\cdot\frac{2v+2}{2v+3}$$

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

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$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nz_n(n-j)(4n+4j-9)(4n-3)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(n-j)(4n+4j-3)(4n-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(5n-5j+3)(4n+4j-9)(4n-3)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(7n-7j+6)(4n+4j-9)(4n-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(11n-11j+9)(4n+4j-9)(4n-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(13n-13j+9)(4n+4j-9)(4n-3)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(17n-17j+12)(4n+4j-9)(4n-3)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(19n-19j+15)(4n+4j-9)(4n-3)=\frac{1}{\pi}$$ - $$\sum_{n=0}^{\infty}u_nz_n(23n-23j+18)(4n+4j-9)(4n-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(29n-29j+21)(4n+4j-9)(4n-3)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(31n-31j+24)(4n+4j-7)(4n-3)=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(5n-5j+2)(4n+4j-7)(4n-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(7n-7j+1)(4n+4j-7)(4n-3)=-\frac{1}{\pi}$$

--

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nz_n(4n^2+27A)(4n+B)=\frac{5+B+8A(3+B)}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-23n^2+27)(4n+3)=\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(\phi n^2+27)(4n-5)=-\frac{16}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(2n^2+27)(4n-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-32n^2+27)(n+\phi)=-\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-24n^2+27)(4n+\phi)=-\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-26n^2+27)(4n-13)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-28n^2+27)(4n-9)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-22n^2+27)(4n+3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-48n^2+27)(-n+2)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(-20n^2+27)(4n-1)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-44n^2+27)(-2n+5)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(4n^2+27\phi^3)(4n-\phi^2)=\frac{5+8\phi-\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(4n^2+27\phi^2)(4n-\phi^2)=\frac{13-\phi^2}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(32n^2+27\phi^4)(4n-\phi^2)=\frac{40}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(4n-\phi^2)=\frac{8}{27\phi^2\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(16n^2-27\phi^4)(2n-\phi)=\frac{10}{\pi}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(2An^2+3Bn-C)(4n-3)=\frac{A+B}{\pi}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(2\phi^2 n^2-3\phi n-3)(4n-3)=\frac{1}{\pi}$$

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

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

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

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

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

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+12n-9)(8n-9)=\frac{9}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(An^2+Bn-C)(Dn-9E)=\frac{-32C(D-12E)+36B(D-8E)+9A(5D-36E)}{144\pi}$$

--

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(2n^2-3)(10n-3)=\frac{3}{4\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-20n^3-3n^2+27)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-52n^3-123n^2-300n+387)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(192n^3+20n^2-108n-135)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+4n-9)(8n-9)=\frac{9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(10n^2+10n-27)(20n-27)=\frac{81}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+8n-9)(8n-9)=\frac{9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(10n^2+20n-39)(20n-39)=\frac{13^2}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+\phi n-9)(8n-9)=\frac{9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+4An-9)(4Bn-9)=\frac{15+4A(B-2)-3B}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+4\phi^3 n-9)(4\phi^4 n-9)=\frac{15+9\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-4\phi^3 n-9)(4\phi^4 n-9)=\frac{15-15\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+4\phi^2 n-9)(4\phi n-9)=\frac{13-\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-4\phi^2 n-9)(4\phi n-9)=\frac{21-\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+4\phi n-9)(4\phi^2 n-9)=\frac{18-\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-4\phi n-9)(4\phi^2 n-9)=\frac{10-\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-4\phi^{-1} n-9)(4\phi^4 n-9)=\frac{3-3\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+4\phi^{-1} n-9)(4\phi^4 n-9)=\frac{27-3\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2+17\phi^2 n-9)(4\phi n-9)=-\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-14\phi n-9)(4\phi^2 n-9)=-\frac{\phi^4}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-5\phi^{-1} n-9)(4\phi^4 n-9)=-\frac{3\phi^4}{\pi}$$

---

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-3)(16n-3)=\frac{9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(7n^2-9)(28n-9)=\frac{27}{2\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(2n^2-15)(8n-15)=\frac{75}{2\pi}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(26n^2-45)(104n-45)=\frac{225}{2\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(7n^2-15)(28n-15)=\frac{25}{3\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-9)(16n-9)=\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(16n^3-48n^2-192n+171)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(292n^3-201n^2-804n+468)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-1796n^3+489n^2+1956n+18)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-380n^3+159n^2+636n-207)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-32n^3+12n^2+48n-9)=\frac{1}{\pi}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(22n^2+3)(44n+3)=\frac{9\cdot 39}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(4n^2-9)(8n-9)=\frac{9}{\pi}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(27-16n^3)=\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(28n^3-15n^2-30n+9)=\frac{1}{\pi}$$

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$ \sum_{n=0}^{\infty}u_nw_n(-87n^2+30n+8)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-180n^3+498n^2-30n-17)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(12n^2+12n+1)(12n+1)=\frac{16}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(38n^2+38n-1)(38n-1)=\frac{139}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(144n^3+21n^2-66n-14)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(-144n^3-3n^2+78n+16)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(-1444n^3-146n^2+916n+139)=\frac{1}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(1444n^3+164n^2-904n-137)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(44n^2-1)=\frac{4}{3\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(-212n^3+1130n^2-352n-45)=\frac{1}{\pi}$$

---

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(34n^2-44n+1)=\frac{1}{\pi}$$

--

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(34n^2+34n+9)(34n+9)=\frac{9^2\cdot5}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(58464n^4-1156n^3-133006n^2-29844n+43767)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(n^2+n-3)(2n-3)=\frac{9}{10\pi}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}$$

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

$$\sum_{n=0}^{\infty}u_nw_nn=\frac{4\pi}{\Gamma^4\left(\frac{1}{4}\right)}$$

---

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nw_n(28n^2-3n-3)(28n-3)=\frac{3^2\cdot21}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(316n^2+45n+45)(316n+45)=\frac{45^2\cdot21}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(128n^2+9n+9)(128n+9)=\frac{9^2\cdot74}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nw_n(248n^2-15n-15)(248n-15)=\frac{15^2\cdot74}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(16n^2+16n+3)(16n+3)=\frac{180}{\pi}$$

--

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nw_n=\frac{(2v)!!^2}{(2v+1)!!^3}(2v+5)\cdot\frac{2^{v-2}}{\pi}$$

$$ \sum_{n=0}^{\infty}u_nz_n=\frac{(2v)!!^2}{(2v+1)!!^3}(2v+5)\cdot\frac{2^{v-1}}{\pi}$$

---

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nw_n(23912n^2-12516n-37053)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(1909352n^2-981540n-2992653)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(156968n^2-81204n-245061)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(22176n^3-22660n^2-27852n+16623)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(280080n^3-286480n^2-351540n+209979)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(32n-3)^2=\frac{6^3}{\pi}$$

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

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

$$\sum_{n=0}^{\infty}u_nv_n(28n+3)^2=\frac{38\cdot9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(80n-3)^2=\frac{252\cdot9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(6n+6)^2=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(22n+3)^2=\frac{9\cdot17}{\pi}$$

$$\sum_{n=0}^{\infty}u_nz_n(80n+9)^2=\frac{9^2\cdot24}{\pi}$$

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nv_n\left[(30j-22)n+1\right]^2=\frac{89j^2-126j+45}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n\left[(30j-8)n-1\right]^2=\frac{89j^2-52j+8}{\pi}$$

-

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n(36n^3-12n^2+6\phi^3n-\phi^2-1)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(72n^3-33n^2+6(2\phi^2-3)n-\phi^2-3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n\left[9n^2+6(\phi^3+2)n-\phi\right]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n\left[-144n^3+181n^2-3\phi(\phi^4+11)n+2(\phi^3+2)\right]=\frac{1}{\pi}$$

---

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n(22n-1)(11n-1)=\frac{17}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(28n+1)(14n+1)=\frac{32}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(44n-1)(22n-1)=\frac{72}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(46n+1)(23n+1)=\frac{85}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(62n-1)(31n-1)=\frac{145}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(68n+1)(34n+1)=\frac{184}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(94n-1)(47n-1)=\frac{337}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(92n^2-108n+1)=\frac{2}{\pi}$$

-

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nv_n(-144n^3+181n^2+51n+1)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(-144n^3+181n^2+27n+3)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(216n^3-205n^2-15n-7)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(-216n^3+223n^2+3n+11)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-162n^3+187n^2+12n+5)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-18n^3+15n^2+3n+2)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(-18n^3+15n^2+15n+1)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n(7n^3+31n^2-n+5)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(97n^2-9n-6)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(36n^3-12n^2-6n-1)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(144n^3-75n^2-30n-8)=\frac{1}{\pi}$$

--

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)^4[in-1]^4-[jn-1]^4\right]=F(j,i)$$

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)^3[in-1]^3-[jn-1]^3\right]=G(j,i)$$

$$G(6,3)=-\frac{\pi}{7}$$

$$\sum_{n=0}^{\infty}u_nv_n(an+b)^z=F_z(a,b)$$

$$\sum_{n=0}^{\infty}u_nv_n(3j+1)^2=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(13j+1)(13j+2)=\frac{9}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(6j+1)(6j^2+1)=\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(12j^2-1)=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j+1)(12j^2-1)=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(12j^2+1)=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(18j-2)(18j^2+3)=\frac{17}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(24j^2+3)=\frac{16}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(12j^2+3)=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(3\cdot2^{i}j^2+2i-1)=\frac{2^{i+1}}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(3\cdot2^{i}j^2+3)=\frac{2^{i+1}}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n(12j-1)(3j^2+\phi)=\frac{2}{\pi}$$

---

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)^3[(2i-1)n-1]^3-[(2j-1)n-1]^3\right]=F(j,i)$$

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)^4[(2i-1)n-1]^4-[(2j-1)n-1]^4\right]=G(j,i)$$

--

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)^3[(2i-1)n-1]^3-[(2j-1)n-1]^3\right]=F(j,i)$$

$$F(2j,j)=\frac{-80j^2+108j-105}{12\pi}$$

$$F(3j,j)=\frac{-60j^2+72j-65}{2\pi}$$

$$F(4j,j)=\frac{-320j^2+360j-315}{4\pi}$$

$$F(5j,j)=\frac{-500j^2+540j-465}{3\pi}$$

$$F(6j,j)=\frac{-1200j^2+1260j-1075}{4\pi}$$

$$F(zj,j)=\frac{-20z^2(z-1)j^2+18z(z-1)(z+1)j-15(z^3-1)}{12\pi}$$

$$F(j,zj)=\frac{20z^2(z-1)j^2-18z(z-1)(z+1)j+15(z^3-1)}{12z^3\pi}$$

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)[(2i-1)n-1]-[(2j-1)n-1]\right]=F(j,i)$$

$$F((a-1)j+b,j)=-\frac{8(aj+b)}{9j\pi}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\left[\left(\frac{j}{i}\right)^2[(2i-1)n-1]^2-[(2j-1)n-1]^2\right]=F(j,i)$$

$$ F(j,2j)=\frac{8j-27}{36\pi}$$

$$ F(j,3j)=\frac{8j-24}{27\pi}$$

$$ F(j,4j)=\frac{16j-45}{48\pi}$$

$$ F(j,5j)=\frac{80j-216}{225\pi}$$

$$ F(j,zj)=\frac{4z(z-1)j-9(z^2-1)}{9z^2\pi}$$

$$F(zj,j)=\frac{-4z(z-1)j+9(z^2-1)}{9\pi}$$

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

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

$$\sum_{n=0}^{\infty}u_nv_n[2(3j-1)n+1]^2=\frac{5j^2-2j+1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[2(3j+1)n-1]^2=\frac{5j^2+2j+1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[4(9j^2+1)n^2-4n+1]=\frac{5j^2+1}{\pi}$$

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

$$\sum_{n=0}^{\infty}u_nv_n[(2n+1)^2+n]=\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[(16n-1)^2+n]=\frac{33}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[(18n-1)^2+n]=\frac{42}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[(18n+1)^2+n]=\frac{50}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[(20n+1)^2+n]=\frac{61}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[(34n-1)^2+n]=\frac{154}{\pi}$$

-

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nv_n(17n-1)=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[4jn^2+(9-5j)n]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n[4jn^2-(7+5j)n+2]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nw_n[2(2j-1)n^2-(5j+1)n+1]=\frac{1}{\pi}$$

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

$$\sum_{n=0}^{\infty}u_nv_n\frac{4jn^2+(1-5j)n+1}{(2n-1)^2(n+1)^2}=\frac{1}{\pi}$$

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

---

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

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

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

$$\sum_{n=0}^{\infty}u_nv_n\prod_{j=0}^{v}\frac{1}{n+j+1}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^2\cdot\frac{4v+5}{(4v+6)^2}\cdot\frac{1}{v!}\cdot\frac{1}{\pi}$$

Example: $$v=1$$,

$$\sum_{n=0}^{\infty}u_nv_n=\frac{1}{25\pi}$$

-

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\prod_{j=1}^{v}\frac{1}{n+j}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^2\cdot\frac{1}{v!}\cdot\frac{1}{4\pi}$$

Example: $$v=0$$,

$$\sum_{n=0}^{\infty}u_nv_n=\frac{1}{4\pi}$$

--

$$u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\prod_{j=1}^{v}\frac{1}{n+j}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^2\cdot\frac{4v+5}{4v!}\cdot\frac{1}{\pi}$$

Examples: $$v=0,1$$

$$\sum_{n=0}^{\infty}u_nv_n=\frac{5}{4\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n\frac{1}{n+1}=\frac{1}{\pi}$$

$$ u_n=\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}u_nv_n\prod_{j=1}^{v}\frac{1}{n+j}=\left(\frac{(2v)!!}{(2v+1)!!}\right)^2\cdot\frac{v+1}{v!}\cdot\frac{1}{\pi}$$

Examples: $$v=0,1$$

$$\sum_{n=0}^{\infty}u_nv_n=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}u_nv_n\frac{1}{n+1}=\frac{8}{9\pi}$$

---

$$u_n=\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}$$

$$w_n=\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{10}\right)_n\left(\frac{9}{10}\right)_n}{(1)_n^3}$$

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

$$\sum_{n=0}^{\infty}\frac{u_nv_n}{n+1}=\frac{2^{2v-1}\cdot3^{2(v+1)}[6(v+1)+1]}{(6^2-1)(12^2-1)\cdots[(6(v+1))^2-1]}\cdot\frac{v!}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{w_nv_n}{n+1}=\frac{2^{2v-1}\cdot5^{2(v+1)}[10(v+1)+1]}{(10^2-1)(20^2-1)\cdots[(10(v+1))^2-1]}\cdot\frac{v!}{\phi\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}\cdot\frac{1}{n+1}\prod_{j=0}^{v}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{2^{2v-1}\cdot3^{2(v+1)}[6(v+1)+1]}{(6^2-1)(12^2-1)\cdots[(6(v+1))^2-1]}\cdot\frac{v!}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}\cdot\frac{1}{n+1}\prod_{j=0}^{v}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=\frac{2^{2v-1}\cdot3^{2(v+1)}[6(v+1)+1]}{(6^2-1)(12^2-1)\cdots[(6(v+1))^2-1]}\cdot\frac{v!}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}\cdot\frac{1}{n+1}\prod_{j=0}^{v}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(-4k^2+2k-1)}{(2k-1)(2k+1)}=\frac{2^{2v-1}\cdot3^{2(v+1)}[6(v+1)+1]}{(6^2-1)(12^2-1)\cdots[(6(v+1))^2-1]}\cdot\frac{2v!}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(-\frac{1}{6}\right)_n}{(1)_n^3}\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{27}{35\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{9}{10\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=\frac{9}{10\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(-4k^2+2k-1)}{(2k-1)(2k+1)}=\frac{9}{5\pi}$$

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(1)_n^3}\cdot\frac{1}{n+1}\prod_{j=0}^{v}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{(4v+2)!!}{(4v+3)!!}\cdot\frac{2^v}{(2v+1)!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{5}\right)_n\left(\frac{4}{5}\right)_n}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\left(\frac{5}{4}\right)^2\cdot\frac{\sqrt{2-\phi}}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{10}\right)_n\left(\frac{9}{10}\right)_n}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\left(\frac{5}{3}\right)^2\frac{1}{2\phi\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{25n^2+49n+16}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(-4k^2+2k-1)}{(2k-1)(2k+1)}=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{55n^2+61n-29}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(-4k^2+2k-1)}{(2k-1)(2k+1)}=-\frac{8}{\pi}$$

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{(5n+7)(n-1)}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=-\frac{4}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{13n+17}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{32}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{25n^2+49n+16}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{55n^2+61n-29}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=-\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{25n^2+49n+16}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{23n^2+17n-22}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=-\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{23n^2+17n-22}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=-\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{23n^2+17n-22}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(4k^2-2k+1)}{(2k-1)(2k+1)}=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{n^2+16n+19}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{n^2+16n+19}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{n^2+16n+19}{(n+1)^3(n+2)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)(4k^2-2k+1)}{(2k-1)(2k+1)}=-\frac{8}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)[4\phi k^2+k-\phi^2]}{(2k-1)(2k+1)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(n+1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)[4(j-1)k^2+k-j]}{(2k-1)(2k+1)}=\frac{1}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{n+1}\prod_{j=0}^{v}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{1}{v!}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{n+1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{k(n-k)(1-4k)}{(2k-1)(2k+1)}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{1}{\pi}$$

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

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

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{3n-11k}{(2k-1)(2k+1)}=\frac{2}{\pi}(C-1)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{4n}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{3n-11k}{(2k-1)(2k+1)}=-\frac{1}{\pi}$$

-

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=-\frac{2}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=\frac{1}{\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}(2v+2)\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=-\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\prod_{j=0}^{v}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!}{(2v+1)!!^2}\cdot\frac{(-1)^{v+1}}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\prod_{j=0}^{v}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!}{(2v+1)!!^2}(2v+2)\cdot\frac{(-1)^{v}}{\pi}$$

---

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\prod_{j=1}^{v}\frac{1}{n+j}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{2v+2}{v!}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\prod_{j=0}^v\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{2}{v!}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j+1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}(4v+2)\cdot\frac{1}{\pi};v\ge1$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{n-j-1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{4v+4}{(2v+3)^2}\cdot\frac{1}{\pi};v\ge1$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n+2j+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{(2v+3)^2-1}{(2n+3)^2}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}(2v+2)\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j+1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}(4v+2)\cdot\frac{1}{\pi};v\ge1$$

-

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n+2j+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n+k+1)(n+k+2)}{k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\frac{2v+2}{(2n+3)^2}\cdot\frac{1}{\pi}$$

---

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(8n+8k+5)(16n+16k+5)}{k+1}=\frac{128}{\pi}$$

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

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

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

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

-

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

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{\left(n+k+\frac{5}{4}\right)^2}{k+1}=\frac{1}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{(2n1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{\left(n-\frac{1}{4}k\right)^2}{k+1}=\frac{1}{\pi}$$

---

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{2n-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{(2k+1)(2k+3)}=\frac{\pi}{16}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{(2k+1)(2k+3)}=\frac{\pi}{32}$$

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

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

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(n-k)^2}{k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{6v+5}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{k+1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{4v+4}{\pi}$$

---

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{3n-2k}{(2k-1)(2k-3)}=\frac{C}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{(2k-1)(2k-3)}=\frac{1}{6\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^3}\sum_{k=0}^{n}(-1)^kk{n\choose k}\frac{n-k}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{(2n-2j-1)(2n-2j-3)}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{4v^2+9v+5}{(4v+3)^2}\cdot\frac{1}{\pi}$$

---

$$A=\frac{\pi}{\sqrt{2}\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)},B=\frac{2\sqrt{2}\Gamma^2\left(\frac{3}{8}\right)\Gamma^2\left(\frac{9}{8}\right)}{\pi^3}$$

$$A=2B$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(4n+1)(2n-1)}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=-\frac{\Gamma^4\left(\frac{1}{4}\right)}{32}\cdot\frac{1}{\pi^2}$$

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

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=\frac{(2v+2)!!^2}{(2v+3)!!^2}\cdot\frac{2v+3}{2v+2}\cdot\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{2n-k}{2k-1}=\frac{(2v+2)!!^2}{(2v+3)!!^2}\cdot\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{3n-k}{2k-1}=\frac{(2v+2)!!^2}{(2v+3)!!^2}\cdot\frac{2v+1}{2v+2}\cdot\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{4n-k}{2k-1}=\frac{(2v+2)!!^2}{(2v+3)!!^2}\cdot\frac{2v+0}{2v+2}\cdot\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{2v+2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{2v}{\pi};v\ge1$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{n}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-2k}{2k-1}=\frac{C-1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^3}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-k}{2k-1}=-\frac{2}{\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(2z-1)n-zk}{2k-1}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{(2z+1)n-zk}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{Tn-k}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}\left[2(T-1)v+T-2\right]\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{n-Sk}{2k-1}=\frac{(2v)!!^2}{(2v+1)!!^2}\left[2(S-1)v+2S-1\right]\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{Tn-Sk}{2k-1}=\frac{(2v)!!^2}{(2v+1)!!^2}\left[2(S-T)v+2S-T\right]\cdot\frac{1}{\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}(v+1)\cdot\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j+2}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}(3v+2)\cdot\frac{4}{\pi};v\ge1$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j-1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{2k-1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2n-1}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}v\cdot\frac{4}{\pi}$$

---

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{2k-1}=\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{n+j+1}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{2k-1}=-\frac{(2v)!!^2}{(2v+1)!!^2}\cdot\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{2n+2j+2}{2n-2j-1}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{2k-1}=-\frac{(2v+2)!!^2}{(2v+1)!!^2}\cdot\frac{1}{2v(v+1)}\cdot\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{j=0}^{v}(-1)^j{v \choose j}\frac{1}{(2n-2j-1)(2n-2j-3)}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{2k-1}=-\frac{(2v+2)!!^2}{(2v+3)!!^2}\cdot\frac{1}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{n^3+5n^2}{[(2n-1)(2n-3)(2n-5)]^2}\cdot \frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{2k-1}=-\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{n^3+5n^2}{[(2n-1)(2n-3)(2n-5)]^2}\cdot \frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{k}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{16n^3-n^2}{[(2n-1)(2n-3)(2n-5)]^2}\cdot \frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{k^3}{2k-1}=-\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty} \left(\frac{(n+1)C_n}{4^n}\right)^2\sum_{j=0}^{k}(-1)^j{k \choose j}\frac{1}{n+j+1} =\frac{(2k)!!^2}{(2k+1)!!^2}\cdot \frac{4}{\pi}$$

$$\sum_{n=0}^{\infty} \left(\frac{(2k+1)!!}{(2k)!!}\cdot\frac{(n+1)C_n}{4^n}\right)^2\sum_{j=0}^{k}(-1)^j{k \choose j}\frac{1}{n+j+1} =\frac{4}{\pi}$$

--

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{16^n}\cdot \frac{(n+1)^2}{(n+1)}=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{16^n}\cdot \frac{(n+1)^2}{(n+1)(n+2)}=\frac{16}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{16^n}\cdot \frac{(n+1)^2}{(n+1)(n+2)\cdots(n+j)}=\frac{2^{2j}(j-1)!}{(2j-1)!!^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{16^n}\cdot \frac{(n+1)^2}{(2n-1)}=-\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{16^n}\cdot \frac{(n+1)^2}{(2n-1)(2n-3)}=\frac{4}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{16^n}\cdot \frac{(n+1)^2}{(2n-1)(2n-3)\cdots(2n-2j+1)}=(-1)^j\frac{2^j(j-1)!}{(2j-1)!!^2\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{4^{2n}}\cdot \frac{(n+1)^2}{(2n-1)^2}=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{C_{n}^2}{4^{2n}}\cdot \frac{(n+1)^2}{(2n-1)^2(2n-3)^2}=\frac{32}{\pi}$$

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

-

$$\sum_{n=0}^{\infty}\frac{C_{n-3}^2}{16^n}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+2}=\frac{47}{158760\pi}$$

$$\sum_{n=0}^{\infty}\frac{n+1}{2^{4n-10}}\cdot C_{n-3}^2=\frac{13}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(n+1)(n+2)}{2^{4n}}\cdot (15C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+1)(k+2)(k+3)}=\frac{314}{\pi}$$

-

$$\sum_{n=0}^{\infty}\frac{n+1}{16^{n}}\cdot(3C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+1)(k+2)}=\frac{16}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{n+1}{16^{n+1}}\cdot(15C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+2)(k+3)}=\frac{8}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{n+1}{16^{n+2}}\cdot(105C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+3)(k+4)}=\frac{12}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{n+1}{16^{n+3}}\cdot(945C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+4)(k+5)}=\frac{36}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{n+1}{16^{n+4}}\cdot(10395C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+5)(k+6)}=\frac{180}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{n+1}{16^{n+5}}\cdot(135135C_n)^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(k+6)(k+7)}=\frac{1350}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{1}{(n+1)^2(n+2)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=-\frac{8}{135\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{1}{(n+1)^2(n+2)^2(n+3)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)(2k-7)}=\frac{64}{70875\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{1}{(n+1)^2(n+2)^2\cdots(n+j)^2}\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{(2k-1)(2k-3)\cdots(2k-2j-1)}=(-1)^{j+1}\frac{(2j+1)^2}{j\cdot(2j+1)!!^3}\cdot \frac{2^{2j}}{\pi};j\ge1$$

--

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{1}{(2n-1)^2(2n-3)^2(2n-5)^2\cdots(2n-2j-1)^2}\sum_{k=0}^{n}(-1)^k\frac{2k-1}=-\frac{2^{3j+2}j!}{(2j+1)!!^3}\cdot\frac{1}{\pi};j\ge0$$

-

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

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

--

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

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

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

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

--

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

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

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

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

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

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

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

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

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

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

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(2n-1)(2n-3)(2n-5)\cdots(2n-2j-1)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}=(-1)^{j}\frac{j!2^{j+1}}{(2j+1)!!^2}\cdot\frac{1}{\pi};j\ge0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{n^2}{(2n-1)(2n-3)(2n-5)\cdots(2n-2j-5)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}=(-1)^{j}\frac{(j)!2^{j-1}}{(2j+3)!!^2}\cdot\frac{1}{\pi};j\ge0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}}{(n+1)(n+2)(n+3)\cdots(n+j+1)}=-\frac{j!4^{j+1}}{(2j+1)!!^2}\cdot\frac{1}{\pi};j\ge0$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{8-13n}{(2n-1)(n+2)(n+1)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(n+1)^2(n+2)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=-\frac{2}{15\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(n+1)(n+2)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=-\frac{2}{27\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{(n+1)^2(n+2)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=-\frac{8}{135\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{(2n+1)^2}{(n+1)^3(1-2n)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k+1)}=\frac{4}{9}\left(\frac{4}{\pi}-3\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{(2n+1)}{(n+1)^3(1-2n)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k+1)}=\frac{4}{27}\left(3-\frac{32}{\pi}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{n-1}{(2n+2)^2}\frac{2n+1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k+1)}=\frac{7}{9\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{(2n+1)^2}{(n+1)^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k+1)}=\frac{8}{\pi}-4$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{2n+1}{(n+1)^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k+1)}=4-\frac{16}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{(2n+1)(2n+3)}{(n+1)^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k+1)}=4-\frac{24}{\pi}$$

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

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

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

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

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{(2n-1)(2n+1)}{(n+1)(n+2)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=\frac{1}{5\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{2n-1}{n+2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=\frac{1}{6\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{(2n-1)^2}{(n+1)(n+2)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=-\frac{1}{15\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{n(2n-1)}{(n+1)(n+2)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=\frac{1}{30\pi}$$

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

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

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

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

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

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)_n^3}{(1)_n^3}n^2\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}=\frac{C}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{2}\right)_n^3}{(1)_n^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-3)(2k-5)}=\frac{1}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{1}{(2n-1)(2n-3)\cdots(2n-2j-1)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}=F(j)$$

-

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

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\frac{n^2}{(2n-1)^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}=-\frac{C}{\pi}$$

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

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2\left(-\frac{1}{2}\right)_n}{(1)_n^3}\cdot\frac{n}{2n-1}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}=\frac{1}{\pi} $$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(-\frac{1}{2}\right)_n^2}{(1)_n^3}\cdot\frac{1}{(2n-1)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=\frac{C-1}{2\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{6}\right)_n\left(-\frac{1}{6}\right)_n}{(1)_n^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{k+j}=\frac{6^{2j}\cdot j!}{(6^2-1)(12^2-1)\cdots((6j)^2-1)}\cdot \frac{3}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(-\frac{1}{2}\right)_n}{(1)_n^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{k+1}=\frac{8}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2\left(-\frac{1}{2}\right)_n}{(1)_n^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{k+1}=\frac{\Gamma^4\left(\frac{1}{4}\right)}{3!}\cdot\frac{1}{\pi^3}$$

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2\left(-\frac{1}{2}\right)_n}{(1)_n^3}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{6k-1}{4k^2-1}=\frac{4C}{\pi}$$

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

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

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

$$\sum_{n=0}^{\infty}\frac{H_{n}}{2^n(n+1)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{k+1}=\frac{1}{4}\zeta(3)-\frac{1}{3}\eta^3(1)$$

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

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

$$\sum_{n=0}^{\infty}\frac{H_{n+1}}{2^n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(k+1)^2}=\frac{7}{4}\zeta(3)$$

$$\sum_{n=0}^{\infty}\frac{H_{n+1}^2}{2^n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{k+1}=\frac{7}{4}\zeta(3)$$

$$\sum_{n=0}^{\infty}\frac{H_{n+1}}{2^n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k+1}=2C+\frac{\pi^2}{8}-\frac{\pi}{2}\eta(1)$$

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

$$\sum_{n=0}^{\infty}\frac{H_{n+1}}{2^n(n+1)}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{k+1}=2\zeta(3)-\zeta(2)\eta(1)$$

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

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

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

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

$$\sum_{n=0}^{\infty}\frac{1}{(n+1)2^n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k+3}=\frac{\pi^2}{8}+\pi-4$$

$$\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)(n+3)2^n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k+1}=\frac{\pi^2-4}{32}$$

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$$\sum_{n=0}^{\infty}H_{n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{k^s}{k+j}=(-1)^{j+s}\left(\frac{j}{j-1}\right)^s$$

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

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$$\sum_{n=0}^{\infty}H_{n+2}^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+2}=1+\zeta(2)+H_1+1$$

$$\sum_{n=0}^{\infty}H_{n+2}^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+3}=\zeta(2)-3\zeta(3)+H_2+2$$

$$\sum_{n=0}^{\infty}H_{n+2}^2\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+4}=2\zeta(2)-6\zeta(3)+H_3+3$$

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

$$10\zeta(4)=\sum_{n=0}^{\infty}H_{n}^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+2}+\sum_{n=0}^{\infty}H_{n+1}^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+3}$$

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

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

$$\sum_{n=0}^{\infty}H_{n+1}^3\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{1}{k+2}=10\zeta(4)$$

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

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

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

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

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

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

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

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

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$$\sum_{n=0}^{\infty} \frac{\left(\frac{1}{6}\right)_n\left(-\frac{1}{6}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)\cdots(n+z)}=\frac{6^{2z}}{(6^2-1)(12^{2}-1)\cdots((6z)^2-1)}\cdot\frac{3z!}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{1}{5}\right)_n\left(-\frac{1}{5}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)\cdots(n+z)}=\frac{5^{2z}}{(5^2-1)(10^{2}-1)\cdots((5z)^2-1)}\cdot\frac{5z!\sqrt{3-\phi}}{2\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{1}{4}\right)_n\left(-\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)\cdots(n+z)}=\frac{4^{2z}}{(4^2-1)(8^{2}-1)\cdots((4z)^2-1)}\cdot\frac{2z!\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n\left(-\frac{1}{2}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)\cdots(n+z)}=\frac{2^{2z}}{(2^2-1)(4^{2}-1)\cdots((2z)^2-1)}\cdot\frac{2z!}{\pi}$$

-

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n} {(1)_n^2}\cdot\frac{1}{(n+1)(n+2)\cdots(n+2^z)}=F(z)\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)\cdots(n+z)}=g(z)\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{1}{s}\right)_n\left(1-\frac{1}{s}\right)_n}{(1)_n^2}\cdot\frac{2^{-n}}{(n+1)(n+2)\cdots(n+z)}=g(z,s)$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{n+2^z\pm1}=F(z)\cdot\frac{\sqrt{2}}{\pi}$$

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

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)}=\frac{2^7}{7!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)(n+3)}=\frac{2^{12}}{11!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)(n+3)(n+4)}=3\frac{2^{16}}{15!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)}=15\frac{2^{23}}{19!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)}=15\frac{2^{26}}{23!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=0}^{\infty} \frac{\left(\frac{3}{4}\right)_n\left(\frac{1}{4}\right)_n}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)}=45\frac{2^{31}}{27!!}\cdot\frac{\sqrt{2}}{\pi}$$

$$\sum_{n=1}^{\infty}[\zeta(2n+z)-2\zeta(n+z+1)+\zeta(n+z+2)-\zeta(n+z+3)+\zeta(2n+z+5)]=\sum_{j=1}^{z}\zeta(j+1)-\zeta(z+5)-z$$

$$\sum_{n=1}^{\infty}[\zeta(n+1)-\zeta(n+z+1)]=\sum_{j=1}^{z}\zeta(j+1)-z$$

$$\sum_{n=1}^{\infty}(-1)^n[\zeta(n+1)-\zeta(n+z+1)]=-\sum_{j=1}^{z}(-1)^n\zeta(j+1)-1+(-1)^z$$

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

$$\sum_{n=1}^{\infty}[\zeta(n+1)+\zeta(2n+z)-\zeta(2n+z+1)-\zeta(2n+z+2)-\zeta(2n+z+4)+\zeta(n+z+2)-\zeta(n+z+3)+\zeta(n+z+4)]=\zeta(z+2)$$

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

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

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

$$\sum_{n=1}^{\infty}[\zeta(2n+1)+\zeta(2n+2)+\zeta(n+4)-\zeta(2n+4)-\zeta(2n+5)-\zeta(2n+6)]=\zeta(4)+\zeta(5)+\zeta(6)-2\frac{3}{4}$$

$$\sum_{n=1}^{\infty}[\zeta(2n+1)+\zeta(2n+3)+\zeta(n+5)-\zeta(2n+5)-\zeta(2n+6)-\zeta(2n+7)]=\zeta(5)+\zeta(6)+\zeta(7)-2\frac{3}{4}$$

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

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

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

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

$$\sum_{n=1}^{\infty}[\zeta(n+1)+\zeta(n+1)+\zeta(n+4)-\zeta(2n+4)-\zeta(2n+5)-\zeta(2n+6)]=\zeta(2)+\zeta(4)+\zeta(5)+\zeta(6)-2\frac{3}{4}$$

---

$$\sum_{n=1}^{\infty}[\zeta(n+1)+\zeta(n+s)-\zeta(2n+s)-\zeta(2n+1+s)]=\zeta(s+1)$$

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

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

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

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

---

$$ \int_{-1}^{+1}\frac{\cot^{-1}\left(\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}\frac{\mathrm dx}{1+x^2}=\frac{\pi^2}{16}$$

$$\int_{0}^{\infty}5\cdot\frac{1}{1+y^{10}}\mathrm dy=\int_{0}^{\infty}5\cdot\frac{y^8}{1+y^{10}}\mathrm dy=\pi\phi$$

$$\int_{0}^{\infty}5\cdot\frac{y^2}{1+y^{10}}\mathrm dy=\int_{0}^{\infty}5\cdot\frac{y^6}{1+y^{10}}\mathrm dy=\frac{\pi}{\phi}$$

--

$$\int_{-\infty}^{\infty}2\cdot\frac{y^8}{(1+y^2)(1+y^{8})}\mathrm dy=\pi$$

$$\int_{-\infty}^{\infty}2\cdot\frac{1}{(1+y^2)(1+y^4)(1-y^4+y^{8})}\mathrm dy=\pi$$

$$\int_{-\infty}^{\infty}\frac{5}{2}\cdot\frac{1-y^2}{1+y^{10}}\mathrm dy=\pi$$

$$\int_{-\infty}^{\infty}5y^2\cdot\frac{1-y^2}{1+y^{10}}\mathrm dy=\frac{\pi}{\phi^3}$$

$$\int_{-\infty}^{\infty}\frac{5}{2}\cdot\frac{1}{1+y^2+y^4+y^6+y^8}\mathrm dy=\pi\sqrt{3-\phi}$$

$$\int_{-\infty}^{\infty}\frac{5}{2}\cdot\frac{(1+y+y^2)(1+y^3+y^6)}{(1+y^2+y^4+y^6+y^8)^2}\mathrm dy=\pi\sqrt{3-\phi}$$

$$\int_{0}^{\infty}4\cdot\frac{y^4(1+y^4+y^{16})}{(1+y^2)(1+y^4)(1+y^8+y^{16})}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{2-y^2+y^4}{(1-y^2+y^4)^2}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{y^2(3-y^2+y^4)}{(1-y^2+y^4)^2}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{\sqrt{5}y^6}{1-y^2+y^4-y^6+y^8}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{4y^8}{(1+y^2)(1+y^8)}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{10\phi^3y^2(y^2-1)}{(1+y^2)(1+y^2+y^8+y^{12}+y^{16})}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{10\phi^3y^2(y^2-1)}{(1+y^{10})}\mathrm dy=-\pi$$

$$\int_{0}^{\infty}5\cdot\frac{y^{10}(y^6-1)}{(1+y^2)(1+y^{4}+y^8+y^{12}+y^{14})}\mathrm dy=\pi$$

$$\int_{0}^{\infty}12\cdot\frac{y^2}{(1+y^2)(1+y^{6}+y^{12}+y^{18})}\mathrm dy=\pi$$

$$\int_{0}^{3}\frac{\left(\frac{1}{2-2y+y^2}-\frac{1}{1+y^2}\right)^2}{\frac{1}{2-2y+y^2}+\frac{1}{1+y^2}}\mathrm dy=\frac{2\pi}{\phi^6+1}$$

$$\int_{0}^{\infty}\frac{y^2+2}{(y^2-2y+2)^2}\mathrm dy=\frac{\pi}{2}$$

$$\int_{0}^{\infty}\frac{(y^2-2)^2(y^2+2)}{(y^4-2y^3+4y-4)^2}\mathrm dy=\frac{\pi}{2}$$

$$\int_{0}^{\infty}\frac{8y^5-32y^3+32y}{(y^4-2y^3+4y-4)^2}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{7y^6-40y^3+56}{(0.5y^4-y^3+2y-2)^2}\mathrm dy=15\pi$$

$$\int_{0}^{\infty}\frac{2y^6+16}{(y^4-2y^3+4y-4)^2}\mathrm dy=5$$

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$$\int_{0}^{\infty}\frac{-8y^3-3y^2+60y-38}{y^4-2y^3+4y-4}\mathrm dy=4\ln(2)$$

$$\int_{0}^{\infty}\frac{-16y^3+25y^2-4y-14}{y^4-2y^3+4y-4}\mathrm dy=8\ln(2)$$

$$\int_{0}^{\infty}\frac{3y^2+4y-10}{y^4-2y^3+4y-4}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{3y^2-28y+22}{y^4-2y^3+4y-4}\mathrm dy=-\pi$$

$$\int_{0}^{\infty}\frac{5^2y^2-6^2y-1-2^2-3^2}{y^4-2y^3+4y-4}\mathrm dy=2^2\pi$$

$$\int_{0}^{\infty}\frac{y^{10}+y^8+y^6+y^4}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(y^6-2y^4-3y^2)(\phi y^4-2y^2)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(2y^6-y^4-y^2)(y^4+y^2)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(y^6-y^4-y^2+1)(y^4+y^2+2n)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=3(2n+1)\cdot\frac{\pi}{4}$$

$$\int_{0}^{\infty}\frac{(y^6-y^4-y^2+1)(y^4+y^2+4n-1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=3n\pi$$

$$\int_{0}^{\infty}\frac{(y^6-y^4-y^2+1)(y^2+1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{1-y^4}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(1-y^4)^2}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(2y^4+y^2-1)^2}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{1-y^8}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(3y^2-1)(y^2+1)^3}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\pi$$

$$\int_{0}^{\infty}\frac{(ay^2-1)(y^2+1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=(a-3)\cdot\frac{\pi}{2}$$

$$\int_{0}^{\infty}\frac{(ay^2-1)(by^2+1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=(a-3)(b+1)\cdot\frac{\pi}{4}$$

$$\int_{0}^{\infty}\frac{(2\phi y^2-1)(2\phi y^2+1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\frac{\pi}{4}$$

$$\int_{0}^{\infty}\frac{(5\phi y^2-1)(3y^2+1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\phi^5\pi$$

$$\int_{0}^{\infty}\frac{(7y^2-1)(\phi y^2+1)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\phi^2\pi$$

$$\int_{0}^{\infty}\frac{(7y^2-1)(y^2+\phi)}{y^{12}-10y^{10}+37y^8-42y^6+26y^4-8y^2+1}\mathrm dy=\phi^2\pi$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{1/F_{n+2}}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{1/F_{n+4}}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{1/F_{n+3}}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{1/F_{n+5}}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=\frac{1}{3\pi}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{1/F_{n+k}}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{1/F_{n+k+2}}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=\frac{F_{k-1}}{F_kF_{k+1}}\cdot\frac{2}{\pi}$$

---

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{F_{n+2k}}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{F_{n+2k+2}}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=-\frac{2}{\pi}\cdot F_{2k+1}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{F_{n+2k+1}}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{F_{n+2k+3}}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=-\frac{2}{\pi}\cdot F_{2k}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+2}^z}-\frac{\left(\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+4}^z}\right]=\frac{2^z-1}{2^z}\cdot\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+2}^z}-\frac{\left(\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+4}^z}\right]=\frac{2^z+1}{2^z}\cdot \frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{1}{\tan^{-1}\phi+F_{2n+1}^{\phi}\cot^{-1}\phi}-\frac{1}{\tan^{-1}\phi+F_{2n+3}^{\phi}\cot^{-1}\phi}\right]=\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{1}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{1}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=\frac{4}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{1}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{1}{\tan^{-1}\phi+F_{n+2}^{\phi}\cot^{-1}\phi}\right]=\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{1}{\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi}-\frac{1}{\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi}\right]=0$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+2}}-\frac{\left(\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+4}}\right]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\tan^{-1}\phi+F_{2n+1}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{2n+2}}-\frac{\left(\tan^{-1}\phi+F_{2n+3}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{2n+4}}\right]=\frac{2}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+2}}-\frac{\left(\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+4}}\right]=\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+1}F_{n+2}}-\frac{\left(\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi\right)^{-1}}{F_{n+3}F_{n+4}}\right]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\csc^{-1}\phi+F_{n+1}^{\phi}\sec^{-1}\phi\right)^{-1}}{F_{n+1}F_{n+2}}-\frac{\left(\csc^{-1}\phi+F_{n+3}^{\phi}\sec^{-1}\phi\right)^{-1}}{F_{n+3}F_{n+4}}\right]=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\csc^{-1}\phi+F_{n+1}^{\phi}\sec^{-1}\phi\right)^{-1}}{F_{n+1}F_{n+2}}-\frac{\left(\csc^{-1}\phi+F_{n+3}^{\phi}\sec^{-1}\phi\right)^{-1}}{F_{n+3}F_{n+4}}\right]=\frac{3}{\pi}$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\csc^{-1}\phi+F_{2n+1}^{\phi}\sec^{-1}\phi\right)^{-1}}{F_{2n+1}F_{2n+2}}-\frac{\left(\csc^{-1}\phi+F_{2n+3}^{\phi}\sec^{-1}\phi\right)^{-1}}{F_{2n+3}F_{2n+4}}\right]=\frac{2}{\pi}$$

-

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

$$\sum_{n=1}^{\infty}\left[\frac{1}{F_n^yF_{n+1}^z}-\frac{1}{F_{n+2}^yF_{n+3}^z}\right]=\frac{1+2^z}{2^z},y\le-z$$

$$\sum_{n=1}^{\infty}(-1)^n\left[\frac{1}{F_n^yF_{n+1}^z}-\frac{1}{F_{n+2}^yF_{n+3}^z}\right]=\frac{1-2^z}{2^z},y\le-z$$

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

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

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{1}{F_{n+1}F_{n+2}}\left(y+\frac{y_1}{F_{n+1}^{x-1}}\right)^{-z}-\frac{1}{F_{n+3}F_{n+4}}\left(y+\frac{y_1}{F_{n+3}^{x-1}}\right)^{-z}\right]=\frac{1}{2(y+y_1)^{z}}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\tan^{-1}x+F_{n+1}^{\phi}\cot^{-1}x\right)^{-z}}{F_{n+1}F_{n+2}}-\frac{\left(\tan^{-1}x+F_{n+3}^{\phi}\cot^{-1}x\right)^{-z}}{F_{n+3}F_{n+4}}\right]=\frac{1}{2}\cdot\left(\frac{2}{\pi}\right)^{z}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\tan^{-1}\phi+F_{n+1}^{\phi}\cot^{-1}\phi\right)^{-z}}{F_{n+1}F_{n+2}}-\frac{\left(\tan^{-1}\phi+F_{n+3}^{\phi}\cot^{-1}\phi\right)^{-z}}{F_{n+3}F_{n+4}}\right]=\frac{1}{2}\cdot\left(\frac{2}{\pi}\right)^{z}$$

$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\left(\csc^{-1}\phi+F_{n+1}^{\phi}\sec^{-1}\phi\right)^{-z}}{F_{n+1}F_{n+2}}-\frac{\left(\csc^{-1}\phi+F_{n+3}^{\phi}\sec^{-1}\phi\right)^{-z}}{F_{n+3}F_{n+4}}\right]=\frac{1}{2}\cdot\left(\frac{2}{\pi}\right)^{z};z\ge0$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\csc^{-1}\phi+F_{n+1}^{\phi}\sec^{-1}\phi\right)^{-z}}{F_{n+1}F_{n+2}}-\frac{\left(\csc^{-1}\phi+F_{n+3}^{\phi}\sec^{-1}\phi\right)^{-z}}{F_{n+3}F_{n+4}}\right]=\frac{3}{2}\cdot\left(\frac{2}{\pi}\right)^z$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\csc^{-1}\phi+F_{2n+1}^{\phi}\sec^{-1}\phi\right)^{-z}}{F_{2n+1}F_{2n+2}}-\frac{\left(\csc^{-1}\phi+F_{2n+3}^{\phi}\sec^{-1}\phi\right)^{-z}}{F_{2n+3}F_{2n+4}}\right]=\left(\frac{2}{\pi}\right)^z$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\tan^{-1}\phi+F_{2n+1}^{\phi}\cot^{-1}\phi\right)^{-z}}{F_{2n+1}F_{2n+2}}-\frac{\left(\tan^{-1}\phi+F_{2n+3}^{\phi}\cot^{-1}\phi\right)^{-z}}{F_{2n+3}F_{2n+4}}\right]=\left(\frac{2}{\pi}\right)^z$$


 * P(n)=Prime zeta function*

$$\sum_{n=2}^{\infty}\frac{P(kn)}{n}=-\sum_{n=2}^{\infty}\frac{\mu(n)}{n}\ln\zeta(kn)$$

$$\sum_{n=1}^{\infty}(-1)^{n}\frac{P(kn)}{n}=\ln\left(\frac{\zeta(k)}{\zeta(2k)}\right)$$

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

$$\sum_{n=1}^{\infty}(-1)^n\frac{\mu(n)}{n}\ln\left[\frac{\zeta(2kn)}{\zeta(kn)}\right]=P(k)$$

$$\sum_{n=0}^{\infty}\frac{\mu(2n+1)}{2n+1}\ln\left[\frac{\zeta^2[2k(2n+1)]}{\zeta[k(2n+1)]}\right]=2P(k)$$

$$\sum_{n=1}^{\infty}\frac{4^{\omega(n)}}{n^2}=5$$

$$\sum_{n=1}^{\infty}\frac{2^{\omega(2^kn)}}{n^2}=4$$

$$\sum_{n=1}^{\infty}\frac{2^{\omega(3^kn)}}{n^2}=\frac{9}{2}$$

$$\sum_{n=1}^{\infty}\frac{2^{\omega(7^kn)}}{n^2}=\frac{7^2}{2\cdot5}$$

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

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

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2^{\omega(3^k n)}}{n^2}=\frac{9}{10}$$

-

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\mu(n)^{2k-1}}{n^s}=\frac{1}{\zeta(s)}\cdot\frac{2^s+1}{2^s-1}$$

$$\sum_{n=1}^{\infty}\frac{\mu(n)^{2k}}{n^s}=\frac{\zeta(s)}{\zeta(2s)}$$

$$\sum_{n=1}^{\infty}\frac{\mu(n)^{2k-1}}{n^s}=\frac{1}{\zeta(s)}$$

-

$$2\phi(p^n)-\phi(p^{n+1})+2\phi(p^{n+2})=p^{n+2}-(p^2-p+1)(p-2)p^{n-1};p=2,3,5,7,11...$$

$$2\phi(2^n)-\phi(2^{n+1})+2\phi(2^{n+2})=2^{n+2}$$

$$ 2\phi(2^{kn})-\phi(2^{k(n+1)})+2\phi(2^{k(n+2)})=2^{k(n+2)}-(2^{k-1}-1)2^{kn}$$

$$\frac{\phi(z)}{ 2\phi(z)-\phi(z^2)+2\phi(z^3)}=\frac{1}{2z^2-z+2}$$

$$\sum_{j=0}^{k}{k \choose j}\frac{\phi(z^{j+1})}{\phi(z)}=(z+1)^k$$

$$\sum_{j=0}^{k}(-1)^{j+k}{k \choose j}\phi(z^{j+1})=\phi(z)(z-1)^k$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\phi(n)}{(n)^s}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\left(1-\frac{1}{2^s-1}\right)$$

$$\sum_{n=1}^{\infty}\frac{\phi(2n)}{(2n)^s}=\frac{\zeta(s-1)}{(2^s-1)\zeta(s)}$$

$$\sum_{n=1}^{\infty}\frac{\phi(2n-1)}{(2n-1)^s}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\left(1-\frac{1}{2^s-1}\right)$$

$$\sum_{n=1}^{\infty}\frac{\phi(4n)}{(4n)^s}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^{1-s}}{2^s-1}$$

$$\sum_{n=1}^{\infty}\frac{\phi(4n-2)}{(4n-2)^s}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{1}{2^s}\left(1-\frac{1}{2^s-1}\right)$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sigma_0(n)}{n^s}=\frac{\zeta^2(s)}{2^{2s-1}}\cdot\left[1+\sum_{j=1}^{s-2}2^{j+s}\right]$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sigma_1(n)}{n^s}=\zeta(s)\zeta(s-1)\left(1-\frac{2^s+2^{s-1}-1}{2^{2s-2}}\right)$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sigma_2(n)}{n^s}=\zeta(s)\zeta(s-2)\left(1-\frac{2^s+2^{s-2}-1}{2^{2s-3}}\right)$$

$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sigma_k(n)}{n^s}=\zeta(s)\zeta(s-k)\left(1-\frac{2^s+2^{s-k}-1}{2^{2s-k-1}}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(a\right)_n\left(b-a\right)_n}{\left(1\right)_n^2}=(-1)^{b-1}{a-1 \choose b-1};[1\le b\le a]$$

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

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

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

--

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{4}-\frac{j}{2}\right)_n\left(\frac{3}{4}-\frac{j}{2}\right)_n}{\left(1\right)_n\left(\frac{1}{2}\right)_n}\frac{1}{\phi^{4n}}=\frac{1}{\phi^{\frac{2j-1}{2}}}\left(\frac{1+\sqrt{5^{\frac{2j-1}{2}}}}{2}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{4}-\frac{j}{2}\right)_n\left(\frac{3}{4}-\frac{j}{2}\right)_n}{\left(1\right)_n\left(\frac{1}{2}\right)_n}\frac{1}{5^{n-\frac{7}{4}}}=\sqrt{2^{2j-3}}\left(\sqrt{\phi^{1-2j}}+\sqrt{\phi^{2j-1}}\right)$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}+1\right)_n\left(\frac{j-1}{j}-3\right)_n}{\left(1\right)_n^2}=-\frac{j^2}{(j+1)(2j+1)\pi}\sin\left(\frac{\pi}{j}\right);[j=1,3,5,...]$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}+1\right)_n\left(\frac{j-1}{j}-4\right)_n}{\left(1\right)_n^2}=-\frac{2j^3}{(j+1)(2j+1)(3j+1)\pi}\sin\left(\frac{\pi}{j}\right);[j=1,3,5,...]$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}+1\right)_n\left(\frac{j-1}{j}-5\right)_n}{\left(1\right)_n^2}=-\frac{6j^4}{(j+1)(2j+1)(2j-3)(j^2+j+2)\pi}\sin\left(\frac{\pi}{j}\right);[j=1,3,5,...]$$

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

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}+k\right)_n\left(-\frac{1}{j}-k-1\right)_n}{\left(1\right)_n^2}=(-1)^k\frac{j}{(jk+1)\pi}\sin\left(\frac{\pi}{j}\right)$$

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

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}-\frac{1}{2}\right)_n\left(\frac{j-1}{j}-\frac{1}{2}\right)_n}{\left(1\right)_n^2}=\frac{2j}{(j-2)\pi}\cos\left(\frac{\pi}{j}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}-1\right)_n\left(\frac{j-1}{j}-1\right)_n}{\left(1\right)_n^2}=\frac{j^2}{(j-1)\pi}\sin\left(\frac{\pi}{j}\right)$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}-\frac{3}{2}\right)_n\left(\frac{j-1}{j}-\frac{3}{2}\right)_n}{\left(1\right)_n^2}=\frac{16j^3}{\pi F(j)}\cos\left(\frac{\pi}{j}\right)$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}-\frac{1}{2}\right)_n\left(\frac{j-1}{j}-\frac{1}{2}\right)_n}{\left(1\right)_n\left(\frac{1}{2}\right)_n}=\sin\left(\frac{\pi}{j}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}-\frac{1}{2}\right)_n\left(\frac{j-1}{j}-\frac{1}{2}\right)_n}{\left(1\right)_n\left(\frac{1}{2}\right)_n}\cdot\frac{1}{2^n}=\cos\left(\frac{j-2}{4j}\cdot\pi\right)$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{j}-\frac{1}{2}\right)_n\left(\frac{j-1}{j}-\frac{1}{2}\right)_n}{\left(1\right)_n\left(\frac{1}{2}\right)_n}\cdot\frac{1}{z^n}=\cos\left(\frac{j-2}{j}\cdot\sin^{-1}\left(\frac{1}{\sqrt{z}}\right)\right)$$

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

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{7}{10}\right)_n\left(-\frac{3}{10}\right)_n}{\left(1\right)_n^2}=\frac{50}{21}\cdot\frac{\phi}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(-\frac{9}{5}\right)_n\left(-\frac{6}{5}\right)_n}{\left(1\right)_n^2}=\frac{5^4}{3^2\cdot2^3}\cdot\frac{\sqrt{\frac{\sqrt{5}}{\phi}}}{\pi}$$ $$

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

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

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{1}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{(2n-(2k-1))}=\frac{\pi}{2}\cdot\frac{1}{(2k-1)!!}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}-\frac{1}{s}\right)_n\left(\frac{1}{s}-\frac{1}{2}\right)_n}{\left(1\right)_n^2}=\frac{2s}{s-2}\cdot\frac{\cos(\frac{\pi}{s})}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{3}\right)_n\left(-\frac{1}{3}\right)_n}{\left(1\right)_n^2}=\frac{\sqrt{27}}{2\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{4}\right)_n\left(-\frac{1}{4}\right)_n}{\left(1\right)_n^2}=\frac{\sqrt{8}}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{3}{10}\right)_n\left(-\frac{3}{10}\right)_n}{\left(1\right)_n^2}=\frac{5}{3}\cdot\frac{\phi}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{10}\right)_n\left(-\frac{1}{10}\right)_n}{\left(1\right)_n\left(2\right)_n}=\frac{500}{99}\cdot\frac{1}{\phi\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left(\sqrt{\ z}\right)^{2k-1}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\left(\frac{1+z}{1-z}\right)^{n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left(\sqrt{\tanh z}\right)^{2k-1}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{e^{2zn}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{1}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\frac{1}{z^n}=(-1)^{k}\frac{1}{(2k-1)!!}\left(\frac{z-1}{z}\right)^{[2k-1]/2}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{1}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\frac{1}{(-z)^n}=(-1)^{k}\frac{1}{(2k-1)!!}\left(\frac{z+1}{z}\right)^{[2k-1]/2}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{1}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\frac{1}{\phi^{n-2k+1}}=(-1)^{k}\frac{1}{(2k-1)!!}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{(-1)^n-\sqrt{\phi^{2k-1}}}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\phi^{3n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left(\sqrt[4]{5}\right)^{1-2k}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\phi^{2n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left(\sqrt{\ \phi}\right)^{1-2k}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\phi^{3n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left({\ \phi}\right)^{1-2k}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\sqrt{5^n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left(\sqrt{\ \phi^3}\right)^{1-2k}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\phi^{n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n}{\left(1\right)_n}\frac{\left[1-(-1)^n\left(\sqrt{\ \frac{2}{\sqrt{5}}}\right)^{2k-1}\right]}{(2n-1)(2n-3)\cdots(2n-(2k-1))}\cdot\frac{1}{\phi^{6n}}=0$$

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

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

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

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n}{\left(1\right)_n^2}\cdot\frac{1}{2n-1}\times\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n}{\left(1\right)_n^2}\cdot\frac{n}{(2n-1)^2}=\frac{1}{16}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{\left(1\right)_n^2}\cdot\frac{n}{(2n-1)^2}=\frac{1}{\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{\left(1\right)_n^2}\cdot\frac{(2n+1)(2n+3)(2n+5)}{(2n-1)^2(2n-3)^2}=\frac{314}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{\left(1\right)_n^2}\cdot\frac{n(n-1)(n-2)\cdots(n-(2k-2))}{(2n-1)^2(2n-3)^2\cdots(2n-(2k-1))^2}=\frac{1}{(1,6^2,20^2,56^2,...F(k))\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{\left(1\right)_n^2}\frac{1}{[(2n-1)(2n-3)(2n-5)\cdots(2n-(2k-1))]^2}=\frac{2^{3k}}{2\pi}\cdot\frac{(k-1)!}{(2k-1)!!^3}=\frac{(1)_k}{\left(\frac{1}{2}\right)_k^3}\cdot\frac{1}{2\pi k}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{s}\right)_n\left(1-\frac{1}{s}\right)_n}{\left(1\right)_n\left(\frac{1}{2}\right)_n}\cdot\sin^{2n}(z)=\sec (z)\cos\left(z\cdot\frac{s-2}{s}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{\left(1\right)_n^2}\cdot\frac{(6n+1)(1-2n)}{2^{2n}}=0$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2\left(\frac{1}{2}\right)_n}{\left(1\right)_n^2\left(-\frac{1}{2}\right)_n}\cdot\frac{6n+1}{2^{2n}}=0$$

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

$$\sum_{n=0}^{\infty}\frac{\left(1-\frac{1}{s}\right)_n\left(1+\frac{1}{s}\right)_n}{\left(\frac{1}{2}\right)_n\left(1\right)_n}\cdot\frac{1}{z^n}=\frac{1}{1-\frac{1}{z}}\left[\cos\left(\frac{2}{s}\sin^{-1}\left(\frac{1}{z}\right)\right)+\frac{s}{2}\frac{1}{\sqrt{z(1-\frac{1}{z})}}\sin\left(\frac{2}{s}\sin^{-1}\left(\frac{1}{z}\right)\right)\right]$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{\left(\frac{1}{3}\right)_n\left(1\right)_n}\cdot\frac{1}{2^n}=\frac{1}{\sqrt[4]{2}\sqrt{\sqrt{3}-\sqrt{2}}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{s}\right)_n\left(\frac{1}{2}+\frac{1}{s}\right)_n}{\left(\frac{1}{2}\right)_n\left(1\right)_n}\cdot\frac{2}{z^n}=\left(1-\frac{1}{\sqrt{z}}\right)^{1-2\left(s^{-1}+2^{-1}\right)}+\left(1+\frac{1}{\sqrt{z}}\right)^{1-2\left(s^{-1}+2^{-1}\right)}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{\left(\frac{1}{2}\right)_n\left(1\right)_n}\cdot\frac{1}{z^n}=\sqrt{\frac{z}{z-1}}\cos\left(\frac{2}{3}\sin^{-1}\left(\frac{1}{\sqrt{z}}\right)\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{s}\right)_n\left(1-\frac{1}{s}\right)_n}{\left(\frac{1}{2}\right)_n\left(1\right)_n}\cdot\frac{1}{z^n}=\sqrt{\frac{z}{z-1}}\cos\left(\frac{s-2}{s}\sin^{-1}\left(\frac{1}{\sqrt{z}}\right)\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{6}\right)_n\left(\frac{5}{6}\right)_n}{\left(\frac{1}{2}\right)_n\left(1\right)_n}\cdot\frac{1}{2^n}=\sqrt{\frac{3}{2}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{6}\right)_n\left(\frac{1}{3}\right)_n}{\left(\frac{1}{2}\right)_n\left(1\right)_n}\cdot\frac{1}{2^n}=\sqrt[4]{2}\sqrt{\sqrt{12}-\sqrt{8}+\sqrt{6}-\sqrt{4}}\cdot\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{25}{24}\right)}{\Gamma\left(\frac{13}{24}\right)\Gamma\left(\frac{7}{6}\right)}$$

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{\left(1\right)_n(3)_n}=\frac{32}{9\pi}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{a}{b}\right)_n}{\left(2\right)_n}=\frac{b}{b-a},[a<b]$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n}{\left(1\right)_n^2}\cdot\frac{7n^2+14n+3}{(n+1)^2}=12$$

$$\sum_{n=0}^{\infty}\frac{\sum_{j=1}^{k}\left(\frac{1}{j+1}\right)_n}{\left(2\right)_n}=H_k+k,[k\ge1]$$

$$\prod_{n=0}^{\infty}\frac{\left[\sum_{j=0}^{k}\left(\frac{1}{j+2}\right)_n\right]^{(-1)^{j}}}{\left(2\right)_n}=\frac{\pi}{2}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n}{\left(1\right)_{n+\frac{1}{2}}\left(1\right)_n}\cdot\frac{2n+1}{(n+1)(n+2)(n+3)}=\frac{3}{8\sqrt{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n}{\left(1\right)_{n+\frac{1}{2}}\left(1\right)_n}\cdot\frac{2n+1}{(n+1)(n+2)}=\frac{6}{5\sqrt{\pi}}$$

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{2}{3}\right)_n}{\left(1\right)_{n+\frac{1}{2}}\left(2\right)_n}\cdot\frac{8n+7}{2n+3}=\frac{6}{\sqrt{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{2}{3}\right)_n}{\left(1\right)_{n+\frac{1}{2}}\left(1\right)_n}\cdot\frac{8n+7}{(n+1)(2n+3)}=\frac{6}{\sqrt{\pi}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n\left(\frac{2}{3}\right)_n}{\left(1\right)_n^2}\frac{11n+10}{(n+1)^2(n+2)}=6$$

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{n}^2}{\left(1\right)_{n}^2}\cdot \frac{1}{2n-1}=-\frac{2}{\pi}$$

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{n}^2}{\left(1\right)_{n}^2}\cdot \frac{10n^2+13n+2}{(2n+1)(2n+3)(2n+5)}=\frac{1}{\pi}$$

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{n}^2}{\left(1\right)_{n}^2}\cdot \frac{2n^2-5n+1}{(2n+1)(2n-1)^3}=-\frac{4C}{\pi}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{n+\frac{1}{2}}^k}{\left(2\right)_{n}^k}\cdot \frac{1}{(n+1)(n+2)}=\frac{1}{\sqrt{\pi^k}}+\frac{1}{\sqrt{\pi^k}}\sum_{j=1}^{k}(-1)^j\zeta(j+1)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{n+\frac{1}{2}}^k}{\left(2\right)_{n}^k}\cdot \frac{1}{(n+1)^j}=\frac{\zeta(k+j)}{\sqrt{\pi^k}}$$

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$$\sum_{n=0}^{\infty}\frac{\left(1\right)_{3n+\frac{1}{2}}^k}{\left(\frac{3}{2}\right)_{3n+1}^k}=\frac{2^k-1}{6^k}\cdot\sqrt{\pi^k}\zeta(k)$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{n+\frac{1}{2}}^k}{\left(1\right)_{n+1}^k}=\frac{\zeta(k)}{\sqrt{\pi^k}}$$

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

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{5}{2}\right)_{3n+\frac{1}{2}}^k}{\left(3\right)_{3n+1}^k}=2^k\cdot\left(\frac{2}{3}\right)^{2k}\cdot\frac{\zeta(k)}{\sqrt{\pi^k}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{5}{2}\right)_{6n+\frac{1}{2}}^k}{\left(3\right)_{6n+1}^k}=(2^k-1)\cdot\left(\frac{2}{3}\right)^{2k}\cdot\frac{\zeta(k)}{\sqrt{\pi^k}}$$

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$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\cdot\frac{1}{(n+1)(n+2)(n+3)\cdots(n+k)}=\frac{2^{k+1}}{\pi}\cdot\frac{(2k-2)!!}{(2k-1)!!}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\left[\frac{1}{n+1}+\frac{1}{n+2}\right]=\frac{56}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(1)_n^2}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}\right]=\frac{1756}{225\pi}$$

$$\sum_{n=0}^{\infty}\frac{(2n+1)^3}{2^{6n+16}}\frac{\left(\frac{1}{2}\right)_n^3}{(3)_n^3}(107814n^4+768373n^3+2021742n^2+2316600n+968217)=\frac{47}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(2n+1)^3}{2^{2n+10}}\frac{\left(\frac{1}{2}\right)_n^3}{(3)_n^3}(186n^4+1357n^3+3654n^2+4284n+1833)=\frac{7}{\pi}$$

--

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

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(3)_n^2}(13n+17)=\frac{2^9}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(3)_n^2}(17n^2+25n)=\frac{2^6}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(3)_n^2}(13n^2+13n-8)=-\frac{2^6}{3\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(3)_n^2}(13n^2-25)=-\frac{11\cdot2^6}{9\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{(3)_n^2}(13n^3+13n^2+13n+33)=\frac{23\cdot2^4}{3\pi}$$

---

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(2)_n^32^{6n+9}}(42n+47)(2n+1)^3=\frac{16}{\pi}-5$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(3)_n^32^{6n+11}}(42n+89)(2n+1)^3(2n+3)^3=\frac{8192}{\pi}-2607$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(2)_n^32^{2n+5}}(6n+7)(2n+1)^3=\frac{4}{\pi}-1$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(3)_n^32^{2n+8}}(6n+13)(2n+1)^3(2n+3)^3=\frac{128}{\pi}-39$$

$$\sum_{n=0}^{\infty}\frac{\left(1\right)_n^x}{(2)_n^x}=\zeta(x)$$

$$A=\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{(3)_{n-0.5}^3}=\frac{64}{\pi^{3/2}}[7\zeta(3)-8]$$

$$B=\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_{n+0.5}^3}{(3)_{n}^3}=\frac{64}{\pi^{3/2}}[\zeta(3)-1]$$

$$7B-A=\frac{64}{\sqrt{\pi^3}}$$

$$8B-A=\frac{64\zeta(3)}{\sqrt{\pi^3}}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_{n-2}^2}{(3)_{n-2}^2}=\frac{256}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_{n-1}^2}{(3)_{n-1}^2}=256\left(1-\frac{4}{\pi}\right)$$

$$\sum_{n=0}^{\infty}\frac{\left(1\right)_n}{(2)_n^2}=\operatorname{E_i(1)}-\gamma$$

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

$$\sum_{n=0}^{\infty}\frac{\left(1\right)_{4n}^2}{(2)_{4n}^2}=\frac{C}{2}+\frac{\pi^2}{16}$$

---

Ramanujan:

$$\frac{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^32^{6n}}[42n+5]}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^32^{6n}}\left[2^4-\frac{1}{4}\left(\frac{2n+1}{2n+2}\right)^3\right]}=\frac{1}{\pi}$$

Guillera:

$$\frac{\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{10n}}[820n^2+180n+13]}{\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{10n}}\left[2^7+\frac{1}{8}\left(\frac{2n+1}{2n+2}\right)^5\right]}=\frac{1}{\pi^2}$$

Gourevitch:

$$\frac{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^7}{(n!)^72^{6n}}[168n^3+76n^2n+14n+1]}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^7}{(n!)^72^{6n}}\left[2^5-\frac{1}{2}\left(\frac{2n+1}{2n+2}\right)^7\right]}=\frac{1}{\pi^3}$$

Cullen:

$$\frac{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^7\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^92^{12n}}\left[43680n^4+20632n^3+4340n^2+466n+21\right]}{\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^7\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^92^{12n}}\left[2^{11}-\frac{(2n+1)^7(4n+1)(4n+3)}{8(2n+2)^9}\right]}=\frac{1}{\pi^4}$$

Integers:

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^7\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^92^{12n}}\left[4-\frac{1}{2^{12}}\frac{(2n+1)^7(4n+1)(4n+3)}{(2n+2)^9}\right]=4$$

$${\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^7}{(n!)^72^{6n}}\left[1-\frac{1}{2^{6}}\left(\frac{2n+1}{2n+2}\right)^7\right]}=1$$

--

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^32^{2n}}\left[4+\frac{(2n+1)(4n+1)(4n+3)}{2^2(2n+2)^3}\right]=4 $$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^3\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^52^{4n}}\left[4+\frac{(2n+1)^3(4n+1)(4n+3)}{2^4(2n+2)^5}\right]=4 $$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{2n}}\left[1+\frac{1}{2^{2}}\left(\frac{2n+1}{2n+2}\right)^5\right]=1$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{10n}}\left[1+\frac{1}{2^{10}}\left(\frac{2n+1}{2n+2}\right)^5\right]=1$$

-

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{2n}}(20n^2+8n+1)=\frac{8}{\pi^2}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{2n}}\left[20n^2+8n+9+2\left(\frac{2n+1}{2n+2}\right)^5\right]=\frac{8}{\pi^2}+8$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{2n}}\left[1+\frac{1}{2^2}\left(\frac{2n+1}{2n+2}\right)^5\right]=1$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{10n}}(820n^2+180n+13)=\frac{128}{\pi^2}$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{10n}}\left[820n^2+180n+141-\frac{1}{8}\left(\frac{2n+1}{2n+2}\right)^5\right]=\frac{128}{\pi^2}+128$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n^5}{(n!)^52^{10n}}\left[1+\frac{1}{2^{10}}\left(\frac{2n+1}{2n+2}\right)^5\right]=1$$

-

(1):

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^32^{2n}}(20n+3)=\frac{8}{\pi}$$

(2:)

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(2560n^4+7584n^3+7536n^2+2494n-9)}{(n!)^32^{2n+7}(n+1)^3}=\frac{8}{\pi}-3$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(160n^3+432n^2+406n+131)}{(n!)^32^{2n+7}(n+1)^3}=1$$

(3:)

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(4n+1)(320n^3+968n^2+970n+323)}{(n!)^32^{2n+7}(n+1)^3}=\frac{4}{\pi}+1$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(2n+1)(1280n^3+3952n^2+3952n+1301)}{(n!)^32^{2n+7}(n+1)^3}=\frac{8}{\pi}+7$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(4n+3)(160n^3+504n^2+498n+163)}{(n!)^32^{2n+7}(n+1)^3}=\frac{2}{\pi}+3$$

(3:)

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^32^{2n+7}}\left[128+\frac{(2n+1)(4n+1)(4n+3)}{(n+1)^3}\right]=1$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(4n+1)}{(n!)^32^{2n+7}}\left[320+\frac{(2n+1)(4n+3)}{(n+1)^3}\right]=\frac{4}{\pi}+1$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(2n+1)}{(n!)^32^{2n+7}}\left[1280+\frac{7(4n+1)(4n+3)}{(n+1)^3}\right]=\frac{8}{\pi}+7$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n(4n+3)}{(n!)^32^{2n+7}}\left[160+\frac{3(2n+1)(4n+1)}{(n+1)^3}\right]=\frac{2}{\pi}+3$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{4}\right)_n\left(\frac{3}{4}\right)_n}{(n!)^32^{2n+7}}\left[2560n-\frac{3(2n+1)(4n+1)(4n+3)}{(n+1)^3}\right]=\frac{8}{\pi}-3$$

--

---

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

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

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{3}{4}\right)^{2n}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n\left(\frac{2}{3}\right)_n}{(n!)^3}\left[20n-\frac{(2n+1)(3n+1)(3n+2)}{(2n+2)^3}\right]=\frac{16}{\pi\sqrt{3}}-4$$

-

$$\sum_{n=0}^{\infty}\left[\frac{\left(\frac{1}{2}\right)_n}{(1)_n}\cdot\frac{1}{2^{2n+2}}\right]^3\left[84(2n+1)-\left(\frac{2n+1}{2n+2}\right)^3\right]=\frac{1}{\pi}+1$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\frac{1}{2}\right)_n}{(1)_n}\cdot\frac{1}{2^{2n+2}}\right]^3\left[64-\left(\frac{2n+1}{2n+2}\right)^3\right]=1$$

$$\sum_{n=0}^{\infty}\left[\frac{\left(\frac{1}{2}\right)_n}{(1)_n}\cdot\frac{1}{2^{2n+2}}\right]^3\left[2688n+5\left(\frac{2n+1}{2n+2}\right)^3\right]=\frac{16}{\pi}-5$$

---

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2^{2n+1}}\left[3n-\frac{7}{2}+\left(\frac{2n+1}{2n+2}\right)^3\right]=\frac{1}{\pi}-2$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2^{2n+2}}\left[6(2n+1)-\left(\frac{2n+1}{2n+2}\right)^3\right]=\frac{2}{\pi}+1$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{1}{2^{2n+2}}\left[24n+\left(\frac{2n+1}{2n+2}\right)^3\right]=\frac{4}{\pi}-1$$

---

--- $$\int_{0}^{\infty}t^2\sum_{k=1}^{\infty}\operatorname{csch^2(kt\pi)}\mathrm dt=\frac{\zeta(3)}{6\pi}$$ -

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

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{3}{4}\right)^{2n}\frac{\left(\frac{3}{2}\right)_n\left(\frac{4}{3}\right)_n\left(\frac{5}{3}\right)_n}{(n!)^3}\left(\frac{80}{3n+1}+\frac{40}{3n+2}-\frac{80}{2n+1}+\frac{1}{(n+1)^3} \right)=\frac{16}{\pi\sqrt{3}}+16$$

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{3}{4}\right)^{2n}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n\left(\frac{2}{3}\right)_n(160n^4+462n^3+453n^2+147n-2)}{(n+1)^3(n!)^3}=\frac{128}{\pi\sqrt{3}}-32$$

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{3}{4}\right)^{2n}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n\left(\frac{2}{3}\right)_n(50n^3+123n^2+109n+34)}{(n+1)^3(n!)^3\cdot32}=1$$

$$\sum_{n=0}^{\infty}(-1)^n\left(\frac{3}{4}\right)^{2n}\frac{\left(\frac{1}{2}\right)_n\left(\frac{1}{3}\right)_n\left(\frac{2}{3}\right)_n(4n+3)(10n^3+37n^2+39n+14)}{(n+1)^3(n!)^3}=\frac{32}{\pi\sqrt{3}}+32$$

- $$\sum_{n=0}^{\infty}(-1)^n\frac{(2560n^4+7584n^3+7536n^2+2494n-9)\left(\frac{1}{2}\right)_n \left(\frac{1}{4}\right)_n \left(\frac{3}{4}\right)_n}{2^{2n+7}(n!)^3(n+1)^3}=\frac{8}{\pi}-3$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{(160n^3+432n^2+406n+131)\left(\frac{1}{2}\right)_n \left(\frac{1}{4}\right)_n \left(\frac{3}{4}\right)_n}{2^{2n+7}(n!)^3(n+1)^3}=1$$

--

$$\sum_{n=0}^{\infty}\frac{(2n+3)(12n^3+8n^2-12n-9)\left(\frac{1}{2}\right)_n^3}{2^{2n+4}(n!)^3(n+1)^3}=\frac{1}{\pi}-2$$

$$\sum_{n=0}^{\infty}\frac{(2n+1)(48n^3+140n^2+140n+47)\left(\frac{1}{2}\right)_n^3}{2^{2n+5}(n!)^3(n+1)^3}=\frac{2}{\pi}+1$$

$$\sum_{n=0}^{\infty}\frac{(42n+5)\left(\frac{1}{2}\right)_n^3}{2^{6n}(n!)^3}=\frac{16}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{(21504n^4+64552n^3+64572n^2+21534n+5)\left(\frac{1}{2}\right)_n^3}{2^{6n+9}(n!)^3(n+1)^3}=\frac{16}{\pi}-5$$

$$\sum_{n=0}^{\infty}\frac{(6n+7)(84n^2+156n+73)\left(\frac{1}{2}\right)_n^3}{2^{6n+9}(n!)^3(n+1)^3}=1$$

$$\sum_{n=0}^{\infty}\frac{(2n+1)(672n^3+2012n^2+2012n+671)\left(\frac{1}{2}\right)_n^3}{2^{6n+9}(n!)^3(n+1)^3}=\frac{1}{\pi}+1$$

- $$\sum_{n=0}^{\infty}(-1)^n\frac{\left[2^{15}n(820n+180)(n+1)^5-13(2n+1)^5\right]\left(\frac{1}{2}\right)_n^5}{2^{10n+15}(n!^5)(n+1)^5}=\frac{2^7}{\pi^2}-13$$

$$\sum_{n=0}^{\infty}\frac{n(186n^3+565n^2+567n+189)\left(\frac{1}{2}\right)_n^3}{2^{2n}(n!)^3(n+1)^3}=\frac{124}{\pi}-32$$

$$\sum_{n=0}^{\infty}\frac{(192n^4+584n^3+588n^2+198n+1)\left(\frac{1}{2}\right)_n^3}{2^{2n}(n!)^3(n+1)^3}=\frac{128}{\pi}-32$$

$$\sum_{n=0}^{\infty}\frac{(198n^4+603n^3+609n^2+207n+2)\left(\frac{1}{2}\right)_n^3}{2^{2n}(n!)^3(n+1)^3}=\frac{132}{\pi}-32$$

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

$$\sum_{n=0}^{\infty}\frac{42n\left(\frac{1}{2}\right)_n^3}{4^{3n}(n!)^3}+\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^{3n}(n!)^3}\cdot\frac{5}{(2n+1)^3}=\frac{16}{\pi}$$

-- $$\sum_{n=0}^{\infty}\frac{32\left(\frac{1}{2}\right)_n^3}{4^n(n!)^3}-\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^n(n!)^3}\left(\frac{1}{(n+1)^3}\right)=32$$

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

$$\sum_{n=0}^{\infty}\frac{31\left(\frac{1}{2}\right)_n^3}{4^n(n!)^3}+\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^n(n!)^3}\left(\frac{1}{(2n+1)^3}-\frac{1}{(n+1)^3}\right)=32$$

$$\sum_{n=0}^{\infty}\frac{33\left(\frac{1}{2}\right)_n^3}{4^n(n!)^3}-\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^n(n!)^3}\left(\frac{1}{(2n+1)^3}+\frac{1}{(n+1)^3}\right)=32$$

$$\sum_{n=0}^{\infty}\frac{192n(n+1)^3\left(\frac{1}{2}\right)_n^3+\left(\frac{3}{2}\right)_n^3}{(n+1)^34^n(n!)^3}=\frac{128}{\pi}-32$$

$$\sum_{n=0}^{\infty}\frac{6n(2n+1)^3\left(\frac{1}{2}\right)_n^3+\left(\frac{3}{2}\right)_n^3}{(2n+1)^34^n(n!)^3}=\frac{4}{\pi}$$ -

$$\sum_{n=0}^{\infty}\frac{198n\left(\frac{1}{2}\right)_n^3}{4^n(n!)^3}+\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^n(n!)^3}\left(\frac{1}{(n+1)^3}+\frac{1}{(2n+1)^3}\right)=\frac{132}{\pi}-32$$

$$\sum_{n=0}^{\infty}\frac{186n\left(\frac{1}{2}\right)_n^3}{4^n(n!)^3}+\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^n(n!)^3}\left(\frac{1}{(n+1)^3}-\frac{1}{(2n+1)^3}\right)=\frac{124}{\pi}-32$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{3}{2}\right)_n^3}{4^n(n!)^3}\left(\frac{32}{(2n+1)^3}-\frac{1}{(n+1)^3}\right)=32$$

-

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^2}{2^n(n+1)(n!)^2}=\frac{8\sqrt{\pi}}{\Gamma\left(\frac{1}{4}\right)}$$

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

$$\sum_{n=0}^{\infty}\frac{2\left(\frac{1}{2}\right)_n^2-\left(\frac{3}{2}\right)_n^2}{2^n(n+1)^2(n!)^2}=\left(\frac{8}{\Gamma\left(\frac{1}{4}\right)}\right)^2\sqrt{\pi}-8$$

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

--

$$\sqrt{5}\sum_{n=0}^{\infty}\frac{F_{2n}}{2^{2n}}\cdot\frac{\Gamma^2\left(n+\frac{1}{2}\right)}{n+\frac{1}{2}}\cdot\frac{1}{n!\left(\frac{1}{2}\right)_n}=4\pi\left(\phi^{-1}\sin^{-1}\left(\frac{\phi}{2}\right)-\phi\sin^{-1}\left(\frac{\phi^{-1}}{2}\right)\right)$$

$$ \sum_{n=0}^{\infty}\frac{L_{2n}}{2^{2n}}\cdot\frac{\Gamma^2\left(n+\frac{1}{2}\right)}{n+\frac{1}{2}}\cdot\frac{1}{n!\left(\frac{1}{2}\right)_n}=4\pi\left(\phi^{-1}\sin^{-1}\left(\frac{\phi}{2}\right)+\phi\sin^{-1}\left(\frac{\phi^{-1}}{2}\right)\right)$$

$$ \sum_{n=0}^{\infty}(-1)^n\frac{L_{2nj}}{2^{2n}n!\left(\frac{1}{2}\right)_n}=\cos\left(\phi^{j}\right)+\cos\left(\phi^{-j}\right)$$

$$ \sum_{n=0}^{\infty}\frac{L_{2nj}}{2^{2n}n!\left(\frac{1}{2}\right)_n}=\cosh\left(\phi^{j}\right)+\cosh\left(\phi^{-j}\right)$$

$$ \sum_{n=0}^{\infty}(-1)^n\frac{L_{2nj}}{2^{2n}(2n+1)n!\left(\frac{1}{2}\right)_n}=\phi^{-j}\sin\left(\phi^{j}\right)+\phi^{j}\sin\left(\phi^{-j}\right)$$

$$ \sum_{n=0}^{\infty}\frac{L_{2nj}}{2^{2n}(2n+1)n!\left(\frac{1}{2}\right)_n}=\phi^{-j}\sinh\left(\phi^{j}\right)+\phi^{j}\sinh\left(\phi^{-j}\right)$$

--- $$ \sqrt{5}\sum_{n=0}^{\infty}\frac{F_{(2j)n}}{(-4)^{n}(2n+1)n!\left(\frac{1}{2}\right)_n}=\phi^{-j}\sin\left(\phi^{j}\right)-\phi^{j}\sin\left(\phi^{-j}\right)$$

$$ \sqrt{5}\sum_{n=0}^{\infty}\frac{F_{(2j)n}}{2^{2n}(2n+1)n!\left(\frac{1}{2}\right)_n}=\phi^{-j}\sinh\left(\phi^{j}\right)-\phi^{j}\sinh\left(\phi^{-j}\right)$$

$$ \sqrt{5}\sum_{n=0}^{\infty}\frac{F_{n}}{2^{2n}(2n+1)n!\left(\frac{1}{2}\right)_n}=\sqrt{\phi^{-1}}\sinh\left(\sqrt{\phi}\right)-\sqrt{\phi}\sin\left(\sqrt{\phi^{-1}}\right)$$

$$ \sqrt{5}\sum_{n=0}^{\infty}\frac{F_{n}}{(-4)^{n}(2n+1)n!\left(\frac{1}{2}\right)_n}=\sqrt{\phi^{-1}}\sin\left(\sqrt{\phi}\right)-\sqrt{\phi}\sinh\left(\sqrt{\phi^{-1}}\right)$$

$$ \sqrt{5}\sum_{n=0}^{\infty}\frac{F_{(2j)n}}{(-4)^{n}n!\left(\frac{1}{2}\right)_n}=\cos\left(\phi^{j}\right)-\cos\left(\phi^{-j}\right)$$

$$ \sqrt{5}\sum_{n=0}^{\infty}\frac{F_{(2j)n}}{2^{2n}n!\left(\frac{1}{2}\right)_n}=\cosh\left(\phi^{j}\right)-\cos\left(\phi^{-j}\right)$$

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

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

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

$$\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}\frac{H_{j+s}}{j(j+1)(k+1)^t}=\frac{H_s}{s}\cdot(1+s)\left[\zeta(t)-1\right]$$

$$\sum_{k=1}^{\infty}\frac{H_k}{2^k(k+1)}=\ln^2(2)$$

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

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

$$\sum_{k=1}^{\infty}\frac{H_k}{k^4}\left[3kH_k-4(-1)^k-5\right]=\frac{23}{8}\zeta(5)$$

$$\sum_{k=1}^{\infty}\frac{H_k}{k^6}\left[kH_k+4(-1)^k+1\right]=-\frac{57}{32}\zeta(7)$$

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$$ \lim_{n \to \infty}\left[ \frac{2n-1}{2n}-\ln n+\sum_{k=z}^{n} \left(\frac{1}{k}-\frac{\zeta(1-k)}{n^k}\right)\right]=\gamma+1-H_{z-1}$$

$$ \lim_{n \to \infty}\left[ \frac{1}{2n}+H_n+\sum_{k=z}^{n} \left(\frac{\zeta(1-k)}{n^k}-\frac{1}{k}\right)\right]=H_{z-1}$$

$$ \lim_{n \to \infty}\left[ \frac{n-1}{n}-H_n-\ln n+2\sum_{k=z}^{n} \left(\frac{1}{k}-\frac{\zeta(1-k)}{n^k}\right)\right]=\gamma+1-2H_{z-1}$$