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The sum of the series $\displaystyle 1+\frac{\sqrt{2}-1}{2\sqrt{2}}+\frac{3-2\sqrt{2}}{12}+\frac{5\sqrt{2}-7}{24\sqrt{2}}+\frac{17-12\sqrt{2}}{80}+.....\infty$

$\bf{My\; Try::}$Here $3-2\sqrt{2} = (\sqrt{2}-1)^2$ and $5\sqrt{2}-7 = (\sqrt{2}-1)^3$

similaryly $17-12\sqrt{2} = (\sqrt{2}-1)^4\;,$ Now let $(\sqrt{2}-1)=x>0$

So our series convert into $$1+\frac{x^2}{2\sqrt{2}}+\frac{x^4}{12}+\frac{x^6}{24\sqrt{2}}+\frac{x^8}{80}+........$$

Now how can i solve after that, Help required, Thanks

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    $\begingroup$ Try to put $y = \frac{x}{\sqrt{2}}$ to simplify it further to $\sum \frac{y^n}{a_n}$ where $a_n$ are integers. Then try to find a pattern in this integer list: $a_n = 2,6,12,20$ for $n=1,2,3,4$ $\endgroup$ – Winther Jun 9 '16 at 3:26
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    $\begingroup$ What exactly is the formula for the $n$th term of the series? $\endgroup$ – DanielWainfleet Jun 9 '16 at 4:12
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From the suggestion by @Winther, express the series as

$$S = 1+\sum_{n=1}^{\infty} a_n x^n $$

where $a_n = \frac1{n (n+1)}$ and $x=1-\frac1{\sqrt{2}} $. Then

$$\begin{align}S &= 1+\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1}\\ &= 1-\log{(1-x)} - \frac1x\left (-\log{(1-x)}-x \right )\\ &= 2-\left (1-\frac1x \right ) \log{(1-x)}\end{align}$$

Now

$$1-\frac1x = 1-\frac{\sqrt{2}}{\sqrt{2}-1} = -\frac1{\sqrt{2}-1}$$

Thus

$$S=2 - \frac12 (\sqrt{2}+1) \log{2} $$

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