Sum of infinite series with Irrational terms 
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

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

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

