User:Proof2012.1

''An inspiring from Sondow's formula we have work out the following ... '' ---

Sondow's formula

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

Definitions

Let

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

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

$$E=\sum_{n=1}^{\infty}\frac{1}{2^n-1}=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\cdots=1.60669...$$

and

$$E_s=\sum_{n=1}^{\infty}\frac{1}{2^n+1}=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots=0.76449...$$

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Overall summary of this paper

Pair 1

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

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

and

Pair 2

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

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

Pair 3

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

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

Pair 4

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

$$E_s=k-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

Pair 5

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

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

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Random 6

$$\frac{E+E_s}{2}-1=\frac{1}{2(2^2-1)}+\frac{1}{2^2(2^4-1)}+\frac{1}{2^3(2^6-1)}+\cdots$$

$$\frac{3}{2}-2z=E_s\cdot$$

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An inspiring formula from Sondow

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

It is from this formula that give me an idea to write ...

Page 1

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Definition

Let

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

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

I have noticed that the sum of the following fractions give the reciprocal power of 2.

e.g,

$$\frac{1}{2\cdot3}+\frac{1}{3\cdot4}=\frac{1}{4}=\frac{1}{2^2}$$

$$\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=\frac{1}{8}=\frac{1}{2^3}$$

So the above series $$X_2$$ can be rewritten as

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

This series can be evaluated by using Arithmetic-Geometri formula

$$X_2=\sum_{n=1}^{\infty}\frac{n}{2^n}=2$$

or

$$\frac{X_2}{2}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}=1$$

Page 2.

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Definition

Let

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

$$Y_1$$ be the Vacca series,

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

$$Y_2=2\ln2-1=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\cdots$$

Evaluation

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Evaluate $$Y_2-Y_0\cdot$$

$$Y_2=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}+\frac{1}{7\cdot8}-\frac{1}{8\cdot9}+\cdots$$

Expanded $$Y_0\cdot$$

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

$$Y_2-Y_0=0+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{2}{4\cdot5}-\frac{2}{5\cdot6}+\frac{2}{6\cdot7}-\frac{2}{7\cdot8}+\frac{3}{8\cdot9}-\frac{3}{9\cdot10}+\cdots$$

Factorise [$$Y_2-Y_0\cdot$$] and we have

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

Evaluate $$Y_1-[Y_2-Y_0]\cdot$$

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

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

e.g,

$$\frac{1}{2}-\frac{1}{2\cdot3}=\frac{1}{3}$$

$$-\frac{1}{3}+\frac{1}{3\cdot4}=-\frac{1}{4}$$

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

Therefore we have

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

I have noticed that if I added this series [$$Y_1-Y_2+Y_0$$] to $$Y_1$$ I will end up with series $$\frac{X_2}{2}$$ which I have evaluated in page 1.

Evaluate of [$$Y_1-Y_2+Y_0\cdot$$] + [$$Y_1\cdot$$]

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

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

e.g,

$$1\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{2}-\frac{1}{4}$$

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

and so on ...

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

Simplify, we have

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

Where 4, 8, 16, ... are power of 2. Rewrite the series and we have

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

Recall from page 1, where $$X_2\cdot$$

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

Divide 2 by both sides and we have

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

Therefore we have

$$2Y_1-Y_2+Y_0=1\cdot$$

From here we solve for $$Y_0$$

$$Y_0=1-2Y_1+Y_2\cdot$$

Recall from definition, where

$$Y_1=\gamma\cdot$$  and

$$Y_2=2\ln2-1\cdot$$

Substitute these values and we solve for $$Y_0$$

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

Therefore we have,

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

So we have worked out the close-form for $$Y_0$$

Recall

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

and therefore we have

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

'''Let us rewrite this series in a neater form ''' $$2\ln2-2\gamma=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-\cdots$$

Add -1 to both sides

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

Then multiply by -1 to both sides and we have

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

From this series, we can derive some other interesting close-form for other series.

Recall from definition,

$$Y_2=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}+\frac{1}{7\cdot8}-\frac{1}{8\cdot9}+\cdots$$

Let us add these two series [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_2\cdot$$]

Evaluation of [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_2\cdot$$]

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Expand [$$2\gamma+1-2\ln2\cdot$$] and we have

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

[$$2\gamma+1-2\ln2\cdot$$] + [$$Y_2\cdot$$] and we have

$$2\gamma+1-2\ln2+Y_2=1+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{2}{4\cdot5}-\frac{2}{5\cdot6}+\frac{2}{6\cdot7}-\frac{2}{7\cdot8}+\cdots$$

Factorise the terms

$$2\gamma+1-2\ln2+Y_2=1+1\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

Recall $$Y_2=2\ln2-1\cdot$$,

substitute $$Y_2\cdot$$ into [$$2\gamma+1-2\ln2+Y_2\cdot$$] and we have,

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

Recall from definition

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

Let us evaluate [$$Y_1\cdot$$] - [$$2\gamma-1\cdot$$]

e.g,

$$\frac{1}{2}-\frac{1}{2\cdot3}=\frac{1}{3}$$

$$-\frac{1}{3}+\frac{1}{3\cdot4}=-\frac{1}{4}$$, and so on ...

and we have

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

Question to be asked

We know the close-form of the following series.

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

and

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

except

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

Before we move on to the next page we can derive another close-form from series [$$2\gamma+1-2\ln2$$]

Let introduce another new series namely $$Y_a\cdot$$, where

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

Recall [$$2\gamma+1-2\ln2\cdot$$]

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

Evaluate [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_a\cdot$$]

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As for this series $$Y_a$$, I don't remember how I derived it.

Evaluation of [$$2\gamma+1-2\ln2\cdot$$] + [$$Y_a\cdot$$], and we have

$$2\gamma+1-2\ln2+Y_a=1+6\left(\frac{1}{4\cdot5}-\cdot-\frac{1}{7\cdot8}\right)+10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Substitue in the value for $$Y_a\cdot$$

$$2\gamma+1-2\ln2+2\ln2-1=1+6\left(\frac{1}{4\cdot5}-\cdot-\frac{1}{7\cdot8}\right)+10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Simplify

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

Divide 2 by both sides and we have

$$\gamma=\frac{1}{2}+3\left(\frac{1}{4\cdot5}-\cdots-\frac{1}{7\cdot8}\right)+5\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)+7\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)+\cdots$$

Page 3.

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Definition

Let

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

Recall from page 2,

$$Y_2=2\ln2-1=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\cdots$$

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

Evaluation

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As for the series $$Y_2$$ I am going to rewrite it as,

$$2-2\ln2=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\cdots$$

We will called it $$Y_3\cdot$$, so we have

$$Y_3=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\cdots$$

Evaluate $$Y_3-X_E\cdot$$

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Let us expand $$X_E\cdot$$

$$X_E=1-\frac{2}{1\cdot2}+\frac{2}{2\cdot3}-\frac{3}{3\cdot4}+\frac{3}{4\cdot5}- \frac{3}{5\cdot6}+\frac{3}{6\cdot7}-\frac{4}{7\cdot8}+\cdots$$

Recall $$Y_3\cdot$$

$$Y_3=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\cdots$$

Let us evaluate $$Y_3-X_E\cdot$$,

$$Y_3-X_E=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{2}{3\cdot4}-\frac{2}{4\cdot5}+ \frac{2}{5\cdot6}-\frac{2}{6\cdot7}+\frac{3}{7\cdot8}-\cdots$$

Factorise and we have

$$Y_3-X_E=\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right) +2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+3\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

We will called it $$Y_4=Y_3-X_E\cdot$$

Recall

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

Let us evaluate [$$Y_4\cdot$$] + [$$Y_1\cdot$$] and we have

E.g,

$$\frac{1}{2}+\frac{1}{2}=1$$

$$-\frac{1}{3}-\frac{1}{2\cdot3}=-\frac{1}{2}$$

and so on ...

and again we add [$$Y_4+Y_1\cdot$$] to $$Y_1\cdot$$ and we have

$$Y_4+Y_1=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

$$Y_4+2Y_1=1\left(1-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{15}\right)+\cdots$$

Let us expand on the RHS, and we have

$$Y_4+2Y_1=1-\frac{1}{3}+\frac{2}{3}-\frac{2}{7}+\frac{3}{7}-\frac{3}{15}+\cdots$$

Simplify and we have

$$Y_4+2Y_1=1+\frac{1}{3}+\frac{1}{7}+\frac{1}{15}+\cdots$$

Change the denominator to power of 2

$$Y_4+2Y_1=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\frac{1}{2^4-1}+\cdots$$

On the RHS series, we have Erdos-Borwein constant (E), and we have

$$Y_4+2Y_1=E\cdot$$

Now we substiute the values for $$Y_4\cdot$$ and $$Y_1\cdot$$

Where $$Y_4=2-2\ln2-X_E\cdot$$ and $$Y_1=\gamma\cdot$$ and we have

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

Solve for $$X_E\cdot$$

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

So now we have worked out the close-form $$X_E\cdot$$ (sort of!)

Recall from definition

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

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

Tidy this up,

Add -2 and multiply by -1 to both sides we have,

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

But this is not the series that we are interested in ...

Recall

$$Y_4+Y_1=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

This is the one we are interested in.

$$Y_4+Y_1=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

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

and where $$X_E=2-2\ln2+2\gamma-E\cdot$$, substitute this in the above and we have

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

Simplify and we have

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

Page 4.

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Recall from the definition on Page 1

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

$$X_2=2\cdot$$

and on page 2.

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

Let us called the above series $$Y_3\cdot$$, so

$$Y_3=2\gamma+1-2\ln2\cdot$$

Let us evaluate [$$X_2\cdot$$] + [$$X_3\cdot$$], and we have

$$X_2+X_3=1+2\left(\frac{2}{2\cdot3}\right)+3\left(\frac{2}{4\cdot5}+\frac{2}{6\cdot7}\right)+\cdots$$

Divide both sides by 2 and we have

$$\frac{X_2+X_3}{2}=\frac{1}{2}+2\left(\frac{1}{2\cdot3}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{6\cdot7}\right)+\cdots$$

Substitute in $$X_2\cdot$$ and $$X_3\cdot$$

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

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

We can re-write series $$\left[\frac{3}{2}+\gamma-\ln2\right]$$ as

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

Let us evaluate [$$X_2-X_3\cdot$$] and we have

$$X_2-X_3=0+2\left(\frac{2}{3\cdot4}\right)+3\left(\frac{2}{5\cdot6}+\frac{2}{7\cdot8}\right)+\cdots$$

Divide both sides by 2 and we have

$$\frac{X_2-X_3}{2}=2\left(\frac{1}{3\cdot4}\right)+3\left(\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\right)+\cdots$$

Substitute in $$X_2\cdot$$ and $$X_3\cdot$$

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

Simplify and we have,

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

We can re-write series $$\left[\frac{1}{2}-\gamma+\ln2\right]$$ as

$$1-\gamma+\ln2=1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+3\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+\cdots$$

Evaluate $$\left[2+\gamma-\ln2\right]-\left[1-\gamma+\ln2\right]$$ and we have

$$2-\gamma+\ln2=2\left(\frac{1}{3}\right)+3\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}\right)+ 4\left(\frac{1}{9}-\frac{1}{10}+\cdots+\frac{1}{15}\right)+5\left(\frac{1}{11}-\frac{1}{12}+\cdots+\frac{1}{31}\right)+\cdots$$

Page 5.

Recall definition from page 3

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

page 2

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

and

page 1

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

Where,

$$X_E=2\gamma-2\ln2+2-E\cdot$$ ,

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

and

$$X_2=2\cdot$$

We begin be evaluating $$[X_E] + [Y_0]\cdot$$

-

$$X_E+Y_0=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- \left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

Substitute in $$X_E\cdot$$ and $$Y_0\cdot$$

$$2\gamma-2\ln2+2-E+2\ln2-2\gamma=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- \left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

simplify and we have

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

Where $$X_2\cdot$$ we multiply it by -1 to have

$$-X_2=-\frac{1}{2^1}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-\cdots$$

Let use evaluate [$$-X_2\cdot$$]+[$$2-E\cdot$$]

e.g,

$$\frac{3}{2}-\frac{1}{2}=1$$

$$\frac{1}{1\cdot2}-\frac{1}{3\cdot4}+\frac{1}{2^2}=\frac{2}{3}$$

$$\frac{1}{3\cdot4}-\frac{1}{7\cdot8}+\frac{1}{2^3}=\frac{4}{21}$$

and so on ...

$$-X_2+2-E=1-2\left(\frac{2}{3}\right)-3\left(\frac{4}{21}\right)-4\left(\frac{8}{105}\right)-\cdots$$

I have noticed that

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

$$\frac{4}{21}=\frac{4}{(2^2-1)(2^3-1)}$$

and so on ...

Substitute in $$X_2\cdot$$

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

Simplify and we have

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

Tidy it up by add -1 and then multiply by -1 to both sides and we have

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

Page 6.

-

Recall from page 2

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

The result from page 3

We called this series $$Y_E\cdot$$

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

Result from page 2

We call this series $$Y_X\cdot$$

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

and from page 1

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

We rewrite this series as,

$$\frac{X_2}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots$$

We begin by evaluating [$$Y_E-Y_1\cdot$$]

- $$Y_E-Y_1=1\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+\cdots$$

Substittue in $$Y_E\cdot$$ and $$Y_1\cdot$$ and we have

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

and we called this series $$Y_Y\cdot$$

$$Y_Y=E-2\gamma\cdot$$

Now we evaluate $$Y_X\cdot$$ + $$Y_Y\cdot$$

and we have

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

Substiute in $$Y_X\cdot$$ and $$Y_Y\cdot$$

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

Simplify and we have

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

We called this series $$Y_Z\cdot$$

$$Y_Z=E-1\cdot$$

Let us now evaluate $$\left[Y_Z+\frac{X_2}{2}\right]$$

Recall

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

$$Y_Z+\frac{X_2}{2}=\frac{2}{3}+2\left(\frac{4}{21}\right)+3\left(\frac{8}{105}\right)+\cdots$$

Substitute in for $$Y_Z\cdot$$and $$\frac{X_2}{2}\cdot$$ and we have

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

Simplify and we have

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

Page 7

-

Recall from Page 5, the two results

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

Let A = E + 1

and Page 6

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

Let B = E

We begin by evaluating [A] - [B]

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

Substitute in for [A] and [B]

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

Simplify and we have

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

Recall from page 6

$$Y_Z=E-1\cdot$$

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

Expand and collect the same denominators,

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

Simplify and we have

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

Change the denominators to power of 2, and we have

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

Page 8

--

Definition

Let

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

and recall $$X_E\cdot$$ series from page 3

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

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

and recall $$Y_0$$ from Page 2

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

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

and $$Y_a$$ from page 2

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

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

$$P_X=\frac{8}{9}=\frac{2}{2^2}+\frac{4}{2^4}+\frac{6}{2^6}+\cdots$$

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

We begin by evaluating $$X_A\cdot$$+$$X_E\cdot$$

-

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

Divide both sides by 2, we have

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

and evaluating of $$Y_0\cdot$$+$$Y_a\cdot$$

-

$$Y_0+Y_a=1-4\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 8\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-12\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

Divide by 2

$$\frac{Y_0+Y_a}{2}=\frac{1}{2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)- 4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-6\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

Let $$M=\frac{X_A+X_E}{2}\cdot$$ and $$N=\frac{Y_0+Y_a}{2}\cdot$$

Evaluate M + N

--

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

Recall

$$P_X=\frac{8}{9}=\frac{2}{2^2}+\frac{4}{2^4}+\frac{6}{2^6}+\cdots$$

Now we evaluate [M + N] - [$$P_X\cdot$$]

$$M+N-P_X=\frac{3}{2}-2\left(\frac{2}{3}\right)-4\left(\frac{8}{105}\right) -6\left(\frac{32}{1953}\right)-\cdots$$

Tidy it up, add $$-\frac{3}{2}$$ and then multiply by -1 to both sides and we have

$$-M-N+P_X+\frac{3}{2}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Now substitute the values of M, N, $$P_X$$ on to the LHS and we have

$$\frac{-X_A-X_E-Y_a-Y_0}{2}+\frac{8}{9}+\frac{3}{2}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

$$\frac{-X_A-2\gamma+2\ln2-2+E-2\ln2+1-2\ln2+2\gamma}{2}+\frac{8}{9}+\frac{3}{2}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)} +\frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

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

Evaluating $$X_E-X_A\cdot$$

--

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

Divide both sides by 2, we have

$$\frac{X_E-X_A}{2}=-3\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)- 5\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)-7\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

and evaluating of $$Y_0-Y_a\cdot$$

---

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

Divide by 2

$$\frac{Y_0-Y_a}{2}=-3\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)- 5\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)-7\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

Let $$P=\frac{X_E-X_A}{2}\cdot$$ and $$Q=\frac{Y_0-Y_a}{2}\cdot$$

Evaluate P + Q

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

Recall

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

Now we evaluate [P + Q] - [$$P_Y$$]

$$P+Q-P_Y=-3\left(\frac{4}{21}\right)-5\left(\frac{16}{465}\right) -7\left(\frac{64}{8001}\right)-\cdots$$

Tidy it up, multiply by -1 to both sides and we have

$$-P-Q+P_Y=3\left(\frac{4}{21}\right)+5\left(\frac{16}{465}\right) +7\left(\frac{64}{8001}\right)+\cdots$$

Now substitute for the values of P, Q, $$P_Y\cdot$$ on to the LHS and we have

$$\frac{-2\gamma+2\ln2-2+E+X_A-2\ln2+2\gamma+2\ln2-1}{2}+\frac{11}{18}=3\left(\frac{4}{21}\right)+5\left(\frac{16}{465}\right) +7\left(\frac{64}{8001}\right)+\cdots$$

Simplify and we have

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

Page 9

---

Recall from page 8 the two results

Let this series be R

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

and

and let this series be T

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

Evaluate R + T

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

$$\frac{2E-4}{2}+\frac{54}{18}=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)} +\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Simplify and we have

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

This is the result from page 5

Page 10.

-

Show that,

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

and

$$\frac{1}{45}=\frac{2}{(2^1+1)(2^2+1)(2^3+1)}+\frac{2^2}{(2^2+1)(2^3+1)(2^4+1)}+\frac{2^3}{(2^3+1)(2^4+1)(2^5+1)}+\cdots$$

The above two series are found by trial and error method

Page 11.

---

Definition

Let

$$\frac{X_2}{2}=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=1$$

and

Sondow's formula

$$S_2=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

Evaluate $$\left(\frac{X_2}{2}-S_2\right)$$

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

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

simplify and we have

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

Evaluate $$\left(\frac{S_2}{2}-\frac{X_2}{4}\right)$$

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

simplify and we have

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

---

Page 12

-

An inspiring formula from Sondow

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

Let

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

Expand the series

$$z=\frac{1}{1\cdot2\cdot3}+2\left(\frac{1}{2\cdot3\cdot4}+ \frac{1}{3\cdot4\cdot5}\right)+3\left(\frac{1}{4\cdot5\cdot6}+\cdots+ \frac{1}{7\cdot8\cdot9}\right))+\cdots$$

Re-write it as

$$z-\frac{1}{1\cdot2\cdot3}=2\left(\frac{1}{2\cdot3\cdot4}+ \frac{1}{3\cdot4\cdot5}\right)+3\left(\frac{1}{4\cdot5\cdot6}+\cdots+ \frac{1}{7\cdot8\cdot9}\right))+\cdots$$

Recall from Page 1

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

Re-write it as

$$\frac{X_2}{2}-\frac{1}{4}=\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots=\frac{3}{4}$$

Let us evaluate $$\left(\frac{X_2}{2}-\frac{1}{4}\right)-\left(z-\frac{1}{6}\right)$$

Substitute in $$X_2=2$$

$$\frac{3}{4}-z+\frac{1}{6}=2\left(\frac{1}{2^3}-\frac{1}{2\cdot3\cdot4}- \frac{1}{3\cdot4\cdot5}\right)+3\left(\frac{1}{2^4}-\frac{1}{4\cdot5\cdot6}-\cdots-\frac{1}{7\cdot8\cdot9}\right))-\cdots$$

Simplify and we have

$$\frac{11}{12}-z=2\left(\frac{1}{15}\right)+3\left(\frac{2}{45}\right)+ 4\left(\frac{4}{153}\right)+\cdots$$

I have noticed that

$$\frac{2^0}{(2^1+1)(2^2+1)}=\frac{1}{15}$$

$$\frac{2^1}{(2^2+1)(2^3+1)}=\frac{2}{45}$$

$$\frac{2^2}{(2^3+1)(2^4+1)}=\frac{4}{153}$$

and son on ...

Re-write the above series as

$$\frac{11}{12}-z=\frac{2\cdot2^0}{(2^1+1)(2^2+1)}+\frac{3\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{4\cdot2^2}{(2^3+1)(2^4+1)}+\cdots=z_1$$

Let $$z_1 = \frac{11}{12}-z$$

Let us introduce $$z_2$$

$$z_2=\frac{2^0}{(2^1+1)(2^2+1)}+\frac{2^1}{(2^2+1)(2^3+1)}+ \frac{2^2}{(2^3+1)(2^4+1)}+\cdots=\frac{1}{6}$$

[The series $$z_2$$ I found this through trial and error method, but I have verify it.]

Evaluate $$z_1-z_2$$

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Recall

$$z_1=\frac{2\cdot2^0}{(2^1+1)(2^2+1)}+\frac{3\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{4\cdot2^2}{(2^3+1)(2^4+1)}+\cdots$$

$$z_1-z_2=\frac{1\cdot2^0}{(2^1+1)(2^2+1)}+\frac{2\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{2^2}{(2^3+1)(2^4+1)}+\cdots$$

Substitute in for the values of $$z_1$$ and $$z_2$$

$$\frac{3}{4}-z=\frac{1\cdot2^0}{(2^1+1)(2^2+1)}+\frac{2\cdot2^1}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^2}{(2^3+1)(2^4+1)}+\cdots$$

Multiply 2 by both sides and we have

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

Nice pair

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Let

$$E_s=\sum_{n=1}^{\infty}\frac{1}{2^n+1}=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots=0.7644...$$

Let $$E_s=\frac{3}{2}-2z$$

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

Recall from Page 6

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

[Therefore we have a nice pair!]

A quick summary of Page 12

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Definition

Where E is Erdos-Borwein constant and $$E_s$$ is ? constant.

$$E=\sum_{n=1}^{\infty}\frac{1}{2^n-1}=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\cdots=1.60669...$$

$$E_s=\sum_{n=1}^{\infty}\frac{1}{2^n+1}=\frac{1}{2^1+1}+\frac{1}{2^2+1}+\frac{1}{2^3+1}+\cdots=0.76449...$$

We have

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

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

and

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

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

Continuing from Page 12

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Let

Recall from Page 2

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

New definition ($$U_s$$)

$$U_s=1\left(\frac{1}{4}-\frac{1}{5}\right)+2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+3\left(\frac{1}{10}-\cdots-\frac{1}{17}\right)+\cdots$$

Evaluate $$[1-\gamma]+U_s$$

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$$1-\gamma+U_s=1\left(\frac{1}{3}-\frac{1}{5}\right)+2\left(\frac{1}{5}-\frac{1}{9}\right)+ 3\left(\frac{1}{9}-\frac{1}{17}\right)+\cdots$$

$$1-\gamma+U_s=\frac{1}{3}-\frac{1}{5}+\frac{2}{5}-\frac{2}{9}+\frac{3}{9}-\frac{3}{17}+\cdots$$

$$1-\gamma+U_s=\frac{1}{3}+\frac{1}{5}+\frac{1}{9}+\frac{1}{17}+\cdots$$

Where

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

$$\frac{1}{5}=\frac{1}{2^2+1}$$

$$\frac{1}{9}=\frac{1}{2^3+1}$$

and so on ...

Re-write it as

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

Recall from Definition from Page 12 (summary)

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

Therefore we have

$$1-\gamma+U_s=E_s\cdot$$

Make $$U_s\cdot$$ the subject

$$U_s=E_s+\gamma-1\cdot$$

Evaluation

Re-write $$U_s=E_s+\gamma-1$$ as $$U_s=E_s+\gamma-\frac{3}{3}$$

Recall from page 12 (summary)

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

Multiply by 3 and we have

$$\frac{3}{3}=\frac{3\cdot2^1}{(2^1+1)(2^2+1)}+\frac{3\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{3\cdot2^3}{(2^3+1)(2^4+1)}+\cdots$$ Recall from page 12 (summary)

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

Let us evaluate $$E_s-\frac{3}{3}$$ and we have

$$E_s-\frac{3}{3}=\frac{(1-3)\cdot2^1}{(2^1+1)(2^2+1)}+\frac{(2-3)\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{(3-3)\cdot2^3}{(2^3+1)(2^4+1)}+\frac{(4-3)\cdot2^4}{(2^4+1)(2^5+1)}\cdots$$

Let $$z_3=E_s-\frac{3}{3}$$

$$z_3=\frac{(1-3)\cdot2^1}{(2^1+1)(2^2+1)}+\frac{(2-3)\cdot2^2}{(2^2+1)(2^3+1)}+ \frac{(3-3)\cdot2^3}{(2^3+1)(2^4+1)}+\frac{(4-3)\cdot2^4}{(2^4+1)(2^5+1)}\cdots$$

$$U_s=\gamma+z_3$$

Recall $$U_s$$

$$U_s=1\left(\frac{1}{4}-\frac{1}{5}\right)+2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+3\left(\frac{1}{10}-\cdots-\frac{1}{17}\right)+\cdots$$

Finally we have

$$\gamma+z_3=1\left(\frac{1}{4}-\frac{1}{5}\right)+2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+3\left(\frac{1}{10}-\cdots-\frac{1}{17}\right)+\cdots$$

Continuing from Page 12

Recall from definitions

Let

$$T_1=E$$ and  $$T_2=E_s$$

$$T_1=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

$$T_2=\sum_{n=1}^{\infty}\frac{1}{2^n+1}$$

We have calculated the following series by trial and error method

Let $$T_3=E$$

$$T_3=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-k$$

and $$T_4=E_s$$

$$T_4=k-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

Here we will show that our $$T_3$$ and $$T_4$$ are correct.

Let us show that $$T_1+T_2=T_3+T_4$$

Evaluate of $$T_1+T_2$$

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$$T_1+T_2=\sum_{n=1}^{\infty}\frac{1}{2^n-1}+\sum_{n=1}^{\infty}\frac{1}{2^n+1}$$

simplify and we have

$$T_1+T_2=\sum_{n=1}^{\infty}\frac{2^{n+1}}{2^{2n}-1}$$

Evaluate of $$T_3+T_4$$

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$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-k+k-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

simplify

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n-1}-\sum_{n=1}^{k\to\infty}\frac{2^n}{2^n+1}$$

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n(2^n+1-2^n+1)}{(2^n-1)(2^n+1)}$$

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^n(2)}{2^{2n}-1}$$

Therefore we have

$$T_3+T_4=\sum_{n=1}^{k\to\infty}\frac{2^{n+1}}{2^{2n}-1}$$

Finally we have show that

$$T_1+T_2=T_3+T_4$$

Page 13

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Recall from Page 7

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

Using trial and error method we found close-form for series $$1-E_s$$

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

Evaluate $$\left[E-1\right]-\left[1-E_s\right]$$

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Simplify and we have

$$\frac{E+E_s}{2}-1=\frac{1}{2(2^2-1)}+\frac{1}{2^2(2^4-1)}+\frac{1}{2^3(2^6-1)}+\cdots$$