User:Sammethod

Definitions

Where E is Erdos-Borwein constant

Problem 7.

-

Show that,

$$\frac{1}{9}=\frac{2}{(2^1-1)(2^2-1)(2^3-1)}+\frac{2^2}{(2^2-1)(2^3-1)(2^4-1)}+\frac{2^3}{(2^3-1)(2^4-1)(2^5-1)}+\cdots$$

I found this series through error and trial.

---

Problem 6.

Here, I am trying to find the close form for series $$X_1$$,

[1]

$$X_1=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+3\left(\frac{1}{3\cdot4}-\frac{1}{ 4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

and through the process end up with these two series as shown below,

[2]

$$\frac{X_1+E+2\ln2-3}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)}+ \frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

[3]

$$\frac{E-2\ln2-1-X_1}{2}+\frac{43}{18}=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+ \frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

and we combined [2] and [3] we have which is already known (From problem 1).

$$\frac{X_1+E+2\ln2-3}{2}+\frac{E-2\ln2-1-X_1}{2}+\frac{11}{18}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$\frac{2E-4}{2}+3=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

therefore we have (from problem 1.)

$$E+1=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Here is how I worked it out, for [2] and [3]

---

Let $$X$$ and $$X_1$$ be the two series as shown below,

$$X=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-3\left(\frac{1}{3\cdot4}-\frac{1}{ 4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)-\cdots$$

$$X_1=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+3\left(\frac{1}{3\cdot4}-\frac{1}{ 4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

and

Let $$Y_0$$ and $$X_1$$ be the two series as shown below,

$$Y_0 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

$$Y_1 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

These two series [A] and [B] it is known

[A]

$$-\frac{25}{18}+\frac{1}{2}=-\frac{2}{2^2}-\frac{4}{2^4}-\frac{6}{2^6}-\cdots$$

[B]

$$-\frac{11}{18}=-\frac{3}{2^3}-\frac{5}{2^5}-\frac{7}{2^7}-\cdots$$

The reason I leaves these series in this form, because it makes the calculating process easy to do.

We also known the following (from solving other problems)

$$X=2\gamma-2\ln2+2-E\cdot$$

$$Y_0=2\ln2-2\gamma\cdot$$

$$Y_1=2\ln2-1\cdot$$

Solution 6 to [3]

-

$$X+X_1=2-4\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-8\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)- 12\left(\frac{1}{31\cdot32}-\cdots-\frac{1}{62\cdot63}\right)-\cdots$$

[C]

$$\frac{X+X_1}{2}=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)- 6\left(\frac{1}{31\cdot32}-\cdots-\frac{1}{62\cdot63}\right)-\cdots$$

________________

$$Y_0+Y_1=1-4\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)-8\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)- 12\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

[D]

$$\frac{Y_0+Y_1}{2}=\frac{1}{2}-2\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)- 6\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

Add [C] + [D], which give us

[E]

$$\frac{X+X_1+Y_0+Y_1}{2}=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)-4\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)- 6\left(\frac{1}{31\cdot32}-\frac{1}{63\cdot64}\right)-\cdots$$

[A]

$$-\frac{25}{18}+\frac{1}{2}=-\frac{2}{2^2}-\frac{4}{2^4}-\frac{6}{2^6}-\cdots$$

Add [E] + [A] and we will have

$$\frac{X+X_1+Y_0+Y_1}{2}-\frac{43}{18}=-2\left(\frac{2}{3}\right)-4\left(\frac{8}{105}\right)-6\left(\frac{32}{1953}\right)-\cdots$$

Simplyfing and we have

$$-\frac{X+X_1+Y_0+Y_1}{2}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+ \frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Substitute in Y_0, X, and Y_1, therefore we have

$$\frac{E-2\ln2-1-X_1}{2}+\frac{43}{18}=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+ \frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Solution 6 to [2]

--

$$X-X_1=-6\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)-10\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)- 14\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

[ C' ]

$$\frac{X-X_1}{2}=-6\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)-10\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)- 14\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

________________

$$Y_0-Y_1=-6\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)-10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)- 14\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

[ D' ]

$$\frac{Y_0-Y_1}{2}=-6\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)-10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)- 14\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

Add [C'] + [D'], which give us

[ E' ]

$$\frac{X-X_1+Y_0-Y_1}{2}=-3\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)-5\left(\frac{1}{16\cdot17}-\frac{1}{31\cdot32}\right)- 7\left(\frac{1}{63\cdot64}-\frac{1}{127\cdot128}\right)-\cdots$$

[B]

$$-\frac{11}{18}=-\frac{3}{2^3}-\frac{5}{2^5}-\frac{7}{2^7}-\cdots$$

Add [E'] + [B] and we will have

$$\frac{X-X_1+Y_0-Y_1}{2}-\frac{11}{18}=-3\left(\frac{4}{21}\right)-5\left(\frac{16}{465}\right)-7\left(\frac{64}{8001}\right)-\cdots$$

Simplyfing and we have

$$-\frac{X-X_1+Y_0-Y_1}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)}+ \frac{7\cdot2^6}{(2^6-1)(2^6-1)}+\cdots$$

Substitute in $$Y_0$$, $$X$$, and $$Y_1$$, therefore we have

$$\frac{X_1+E+2\ln2-3}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)}+ \frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

--

Problem 1.

Show that,

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Solution 1.

---

$$X=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

$$2\gamma-2\ln2+2-X=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}  {7}-\frac{1}{15}\right)+\cdots$$

$$2\gamma-2\ln2+2-X=E\cdot$$

$$X=2\gamma-2\ln2+2-E\cdot$$

$$Y_0 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

$$Y_0=2\ln2-2\gamma\cdot$$

$$2=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$$

---

$$X+Y_0\cdot$$

We have

$$2-E=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

$$-2=-\frac{1}{2}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-\cdots$$

Now add these two together and we have

$$-E=1-2\left(\frac{2}{3}\right)-3\left(\frac{4}{21}\right)-4\left(\frac{8}{105}\right) -\cdots$$

Neaten it up, we have

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Problem 2.

-

Show that,

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Solution 2.

---

[1]

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

[2]

$$E-\gamma=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

[3]

$$2\gamma-1=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\frac{1}{9\cdot10}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

[4]

$$1=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots$$

[2] - [1]

We have

$$E-2\gamma=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+ 3\left(\frac{1}{7\cdot8}-\frac{1}{8\cdot9}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

Now we add this to [3]

We have

$$E-1=\frac{1}{1\cdot2}-\frac{1}{3\cdot4}+2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+3\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Now add this to [4]

We have

$$E=\frac{2}{3}+\frac{4}{21}+\frac{8}{105}+\cdots$$

Neaten it up we have

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Problem 3.

--

Show that,

$$1=\frac{2^1}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+ \frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Solution 3,

-- [1]

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

[2]

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

[2] - [1]

We have

$$1=\frac{2^1}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+ \frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Problem 4.

-

Show that,

$$E-1=\frac{1}{2(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\frac{1}{2^4(2^4-1)}\cdots$$

--

$$E-1=\frac{1}{1\cdot2}-\frac{1}{3\cdot4}+2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+3\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Here we need to expand and collect the terms togther

$$E-1=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{7\cdot8}+\frac{1}{15\cdot16}+\frac{1}{31\cdot32}+\cdots$$

Neaten it up and we have

$$E-1=\frac{1}{2(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\frac{1}{2^4(2^4-1)}\cdots$$

Problem 5.

-

Show that,

$$E-\gamma=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

Solution 5.

- [1]

$$\gamma-2\ln2+2-X=1-\frac{1}{1\cdot2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+\cdots$$

[2]

$$2\gamma-2\ln2+2-X=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{15}\right)+\cdots$$

From [2], we have

$$2\gamma-2\ln2+2-X=E\cdot$$

Substitute X into [1], we have

$$E-\gamma=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

-

$$2E-1=\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{2^n+1}{2^n-1}$$

Proposed problem, find the closed-form of the series Z,

$$Z=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{15}\right)+4\left(\frac{1}{15}-\frac{1}{31}\right)+\cdots$$

and also it is equal to Erdos-Borwein constant

Where Z = Z'

$$Z'=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\frac{1}{2^4-1}+\cdots=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

I encountered it, when trying to solve the above problem.

How did I get from the above series to Erdos-Borwein constant is as show below

(1)

$$1-\gamma=\frac{1}{3}-\frac{1}{4}+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

(2)

$$\gamma-2\ln2+2-X=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

(2) - (1)

and we have

$$Z=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\frac{1}{2^4-1}+\cdots=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

____________________________________________________________________________________________

Definition

(1)

$$X=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

(2)

$$-2\ln2+2=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\cdots$$

(3) Vacca's formula

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

________________

Work out

(3) + (2) - (1)

We have

$$\gamma-2\ln2+2-X=1-\frac{1}{1\cdot2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+\cdots$$

2(3) + (2) - (1)

We have

$$2\gamma-2\ln2+2-X=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{15}\right)+\cdots$$

_____________

Curiosity

1st, closed-form is not known

$$\gamma-2\ln2+2-X=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

2nd, Vacca's formula

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

3rd, known

$$1-\gamma=\frac{1}{3}-\frac{1}{4}+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

I want to find the closed-form for Z, so I can work out the close-form for the 1st.

______________________________________________________________________________

$$\sum_{n=1}^{\infty}\frac{N_1(n)N_0(n)}{n(n+1)}=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+ \frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)+\cdots$$

$$\gamma+\frac{3}{2}-\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{6\cdot7}\right)+\cdots$$

$$\frac{1}{2}-\gamma+\ln2=2\left(\frac{1}{3\cdot4}\right)+3\left(\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\right)+\cdots$$

---

Prove that,

$$\gamma=\frac{1}{1\cdot2}+3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\cdots+\frac{1}{7\cdot8}\right)+5\left(\frac{1}{16\cdot17}-\frac{1}{17\cdot18}+\cdots+\frac{1}{31\cdot32}\right)+\cdots$$

Let

$$2\ln2-1=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\cdots-\frac{1}{7\cdot8}\right)+\cdots$$

and

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

Add both of them and we have

$$\gamma=\frac{1}{1\cdot2}+3\left(\frac{1}{4\cdot5}-+\cdots+\frac{1}{7\cdot8}\right)+5\left(\frac{1}{16\cdot17}-\cdots+\frac{1}{31\cdot32}\right)+ 7\left(\frac{1}{64\cdot65}-\cdots+\frac{1}{127\cdot128}\right)+\cdots$$

Prove that,

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

-

Let

$$Y_0 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

Vacca

$$Y_1=\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

$$Y_2=2\ln2-1=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\cdots$$

$$Y_2-Y_0\cdots$$

$$Y_2-Y_0=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+ 2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

$$Y_1-Y_2+Y_0=\left(\frac{1}{3}-\frac{1}{4}\right)+ 2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+ 3\left(\frac{1}{9}+\cdots-\frac{1}{16}\right)+\cdots$$

$$Y_1-Y_2+Y_0+Y_1=\left(\frac{1}{2}-\frac{1}{4}\right)+ 2\left(\frac{1}{4}-\frac{1}{8}\right)+ 3\left(\frac{1}{8}-\frac{1}{16}\right)+\cdots$$

$$2Y_1-Y_2+Y_0=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots$$

$$2\gamma-2\ln2+1+Y_0=1\cdot$$

$$Y_0=2\ln2-2\gamma\cdot$$

Therefore

$$ 2\ln2-2\gamma= \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

Neaten it up and we have

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

--

Result from above formula

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

and

$$2\ln2-1=\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\cdots$$

add both and we have

$$2\gamma-1=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\frac{1}{9\cdot10}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

Vacca

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

Using Vacca with the above, we obtain

$$1-\gamma=\frac{1}{3}-\frac{1}{4}+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

Prove that,

$$1-\gamma=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\sum_{k=1}^{\infty}\frac{1}{(k+1)k^n}$$

$$\ln\left(\frac{2}{1}\right)=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}\right)$$

$$\ln\left(\frac{3}{2}\right)=\frac{1}{3}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1 2^n\cdot3}\right)$$

Now I combined them to get

$$\ln\left(\frac{3}{1}\right)=\frac{1}{2}+\frac{1}{3}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}\right)$$

$$\ln\left(\frac{5}{1}\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\frac{1}{4^n\cdot5}\right)$$

$$1+\ln\left(\frac{k}{1}\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots{\frac{1}{k}}+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots{\frac{1}{k}}-\ln\left(\frac{k}{1}\right)+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1=\gamma+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1-\gamma= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1-\gamma=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\sum_{k=1}^{\infty}\frac{1}{(k+1)k^n}$$

Now Prove that,

$$\ln\left(\frac{b^{b+1}a!}{a^{a+1}b!}\right)=b-a+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left[\frac{1}{(b-1)^n}+\frac{1}{(b-2)^n}+\cdots+\frac{1}{a^n}\right]$$

Let b = 2 and a = 1, we have

$$\ln\left(\frac{2}{1}\right)=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}\right)$$

--

Prove that,

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=1$$

and

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2(n+1)(2n+1)}=1-\gamma$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)}=X_2$$

$$X_2 = \frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)+... $$

$$X_2 = \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{n}{2^n}$$

Using Arithmetic and geometric formulas to work X_2

$$\lim_{n\to\infty}{X_2}=2$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)}=2$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=1$$

--

Sondow's formula

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}-\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=1-\gamma$$

and we have

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2(n+1)(2n+1)}=1-\gamma$$

Prove that,

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)(2n+2)}=\gamma-\frac{1}{2}$$

-

Using Sondow's Formula

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=\gamma-\frac{1}{2}$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)(2n+2)}=\gamma-\frac{1}{2}$$

Prove that

$$\frac{1}{1-\frac{1}{4-\frac{3}{8-\frac{7}{48-\frac{47}{2208-\cdots}}}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n+1}}\right)=3-\phi$$

Millin series

$$\sum_{n=0}^{\infty}\frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}$$

Euler's continued fraction formula

$$1+\frac{1}{a_0}+\frac{1}{a_0a_1}+\frac{1}{a_0a_1a_2}+...=\frac{1}{1-\frac{1}{1+a_0-\frac{a_0}{1+a_1-\frac{a_1}{1+a_2-\cdots}}}}$$

---

$$\sum_{n=0}^{\infty}\frac{1}{F_{2^n}}=\frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_4}+\frac{1}{F_8}+\frac{1}{F_{16}}+\cdots=\frac{7-\sqrt{5}}{2}$$

$$\frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_4}+\frac{1}{F_8}+\frac{1}{F_{16}}+\cdots=\frac{7-\sqrt{5}}{2}$$

$$\frac{1}{F_2}+\frac{1}{F_4}+\frac{1}{F_8}+\frac{1}{F_{16}}+\cdots=3-\phi$$

Now I used

$$F_{2^n}=\prod_{n=0}^{k-1}L_{2^n}$$

to change the following to Ln

$$\frac{1}{L_1}+\frac{1}{L_1L_2}+\frac{1}{L_1L_2L_4}+\frac{1}{L_1L_2L_4L_8}+\cdots=3-\phi$$

L1 = 1

$$1+\frac{1}{L_2}+\frac{1}{L_2L_4}+\frac{1}{L_2L_4L_8}+\cdots=3-\phi$$

$$\frac{1}{1-\frac{1}{1+L_2-\frac{L_2}{1+L_4-\frac{L_4}{1+L_8-\cdots}}}}=3-\phi$$

$$\frac{1}{1-\frac{1}{4-\frac{3}{8-\frac{7}{48-\cdots}}}}=3-\phi$$

--

Definition

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$

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

--

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

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

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

$$2\ln2-\gamma-2\beta^'(0)=\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)2^{4n}}$$

Where

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

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

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

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

-

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

Date:31\1\12

$$1-\gamma=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2(n+1)(2n+1)}$$

--

Date:24/1/12

Tachiya's algebraic infinite product and its equilvalent continued fraction

n :  0  1  2  3  4  5  6 ...

Fn: 0 1 1 2 3 5 8 13 ...

Ln: 2 1  3  4  7  11 18 ...

$$\frac{1}{1-\frac{1}{4-\frac{3}{8-\frac{7}{48-\frac{47}{2208-\cdots}}}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n+1}}\right)=3-\phi$$

Date:28/1/12

$$3-\frac{9}{8-\frac{7}{48-\frac{47}{2208-\cdots}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}}\right)=\frac{3}{\phi}$$

Date:22/1/12

$$\prod_{n=0}^{k-1}L_{2^n}=F_{2^k}$$

Date:25/1/12

$$\frac{1}{1-\frac{1}{8-\frac{7}{48-\frac{47}{2208-\frac{2207}{4870848-\cdots}}}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$$

Date:26/1/12

$$F_{k+1}=\frac{\phi^k}{\sqrt5}\prod_{n=1}^{\infty}\left(1+\frac{L_k}{L_{2^n+k}}\right)$$

$$L_{k+1}=\phi^k\prod_{n=1}^{\infty}\left(1+\frac{F_k}{F_{2^n+k}}\right)$$

Date:27/1/12

$$F_{2k+2}=\phi^{2k}\prod_{n=1}^{\infty}\left(1+\frac{L_kF_k}{L_{2^n+k}F_{2^n+k}}\right)$$

--

$$\frac{1-\phi^{-4a-4b}}{\left(1-\phi^{-2a}\right)^2} \times\left(\frac{\sqrt{5}F_{a}}{L_{a}}\right)^{(-1)^a}= \prod_{n=1}^{\infty}\left(1+\frac{L_bF_b}{L_{2^na+b}F_{2^na+b}}\right)$$

Date:26/1/12

n greater than 0

$$\frac{L_{2^n}L_{2^n+1}-2}{2L_{2^n+1}-L_{2^n}}=F_{2^n+1}$$

$$L_{2^n}L_{2^n+1}L_{2^n+2}+\left(L_{2^n}-L_{2^n+2}\right)\left(1+L_{2^n+1}^2\right)-4L_{2^n+1}=0$$

$$L_{2^n}F_{2^n+1}L_{2^n+2}-\left(L_{2^n}+L_{2^n+2}\right)\left(1+F_{2^n+1}^2\right)+4F_{2^n+1}=0$$

$$F_{2^n+1}\left(L_{2^n}-L_{2^n+2}\right)\left(L_{2^n+1}^2-1\right)+L_{2^n+1}\left(L_{2^n}+L_{2^n+2}\right)\left(F_{2^n+1}^2-1\right)=0$$

---

$$F_{n+1}=\frac{\phi^n+(-1)^n\phi^{-n-2}}{3-\phi}$$

---

Date:14/1/12

$$\tan{A}\prod_{j=0}^{i}\tan\left(60-3^jA\right)\tan\left(60+3^jA\right)= \tan\left(3^{i+1}A\right)$$

Let

{x = tan(A), y = tan(60-A) and z = tan(60+A)}

$$\frac{1}{x}\left(z-y\right)-\frac{1}{y}\left(x+z\right)+\frac{1}{z}\left(x-y\right)=6$$

$$\frac{(yz)^2+1}{yz}-\frac{(xz)^2+1}{xz}+\frac{(xy)^2+1}{xy}=6$$

Date:27/3/11

where n ≥ 0

$$\ln2=\sum_{k=0}^{\infty}(-1)^k\sum_{j=1}^{2n+1}\frac{(-1)^{j+1}}{(2n+1)k+j}$$

$$\frac{\pi}{4}=\arctan\left(\frac{1}{7}\right)+4\left[\arctan\left(\frac{1}{13}\right) +\arctan\left(\frac{1}{21}\right)\right]+2\left[\arctan\left(\frac{1}{31}\right)+ \arctan\left(\frac{1}{43}\right)+\arctan\left(\frac{1}{57}\right)\right]$$

$$\frac{\pi}{4}=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\arctan \left(\frac{1}{n^2F_{2k-1}+nF_{2k+2}+F_{2k+2}}\right)$$

Date:25/12/11

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

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

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

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

$$\frac{\pi-\sqrt{5}\left[J_3^2(\phi^{-1})-J_3^2(\phi^{-2})\right]}{4}= \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\sum_{k=0}^{\infty}\frac{1}{F_{2k+1}^{2n+1}}$$

Date:26/12/11

$$\frac{\pi}{4}=\arctan\left(\frac{1}{\phi}\right)+ \arctan\left(\frac{1}{\phi^3}\right)$$

$$\frac{\pi}{2}=3\arctan\left(\frac{1}{\phi}\right)- \arctan\left(\frac{1}{\phi^5}\right)$$

$$\arctan\left(\frac{1}{5}\right)=\arctan\left(\frac{1}{\phi}\right)- \arctan\left(\frac{1}{\phi^3}\right)-\arctan\left(\frac{1}{\phi^5}\right)- \arctan\left(\frac{1}{\phi^7}\right)$$

$$\arctan\left(\frac{1}{239}\right)=3\arctan\left(\frac{1}{\phi}\right)- 5\arctan\left(\frac{1}{\phi^3}\right)-4\arctan\left(\frac{1}{\phi^5}\right)- 4\arctan\left(\frac{1}{\phi^7}\right)$$

$$\arctan\left(\frac{1}{239}\right)=2\arctan\left(\frac{1}{\phi}\right)- 3\arctan\left(\frac{1}{\phi^3}\right)-3\arctan\left(\frac{1}{\phi^5}\right)- 4\arctan\left(\frac{1}{\phi^7}\right)$$

Date: 4/1/12

$$\frac{\pi}{2}=\arctan\left(\frac{1}{\phi^{\phi}}\right)+ \arctan\left(\frac{1}{\phi^{-\phi+1}}\right)+ \arctan\left(\frac{1}{\phi^{-\phi+2}}\right)- \arctan\left[\frac{1}{(\phi^{-2\phi}+2\phi^{-2})\phi^{-\phi+3}}\right]$$

Definition
$$\phi=\frac{1+\sqrt{5}}{2}$$

Where

$$F_1=1$$

$$F_2=1$$

$$F_{n+2}=F_n+F_{n+1}$$

Where

$$K_0=1$$

$$K_1=1$$

$$K_{n+2}=K_n+K_{n+1}+F_{n+1}$$

Date:1/12/11

Where

x ≥ 1

$$\left(-1\right)^{x-1}\sum_{n=1}^{x}{(-\phi)}^{x+1-n}F_{x+1-n} =-K_{x-1}\phi^{x+1}+\phi^{x-1}\sum_{n=0}^{x-1}K_{n}$$

Date:10/12/11

$$4\phi^4+3\phi^3+2\phi^2+1\phi=\phi^8$$

$$2\phi^2+1\phi=\phi^4$$

Date: 16/12/11

$$F_m\phi^n-F_n\phi^m=\frac{F_mF_{n+x}-F_nF_{m+x}}{F_x}$$

Date:18/12/11

$$\frac{\phi^n-2F_{n+1}}{\phi^k-2F_{k+1}}= \frac{\phi{F_{n-1}}+F_{n+1}}{\phi{F_{k-1}}+F_{k+1}}$$

Date:23/12/11

Where

n : 0, 1, 2, 3, 4,  5,  6,...

Tn : 0, 1, 3, 6, 10, 15, 21,...

$$\frac{\pi^2}{6}=1-\frac{1}{4T_0+1-\frac{1^4}{4T_1+1-\frac{2^4}{4T_2+1-\frac{3^4}{4T_3+1-...}}}}$$