Evaluating an infinite series $\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$ I have been trying to find the sum of the series $$\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$
but I couldn't find any methods (such as a fourier series) that seem to get me anywhere. 
WolframAlpha gave $ \dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$ but how would one get to this?
 A: Hint Let $f(x)=\sum_{n=1}^\infty \frac{1}{n^2}x^n$. Then 
$$f'(x)=\sum_{n=1}^\infty \frac{1}{n}x^{n-1} \\
xf'(x)=\sum_{n=1}^\infty \frac{1}{n}x^{n}\\
f'(x)+xf''(x)=\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}$$
For simplicity let $y=f'(x)$. Then you need to solve:
$$y+xy'=\frac{1}{1-x}$$
or equivalently 
$$(xy)'=\frac{1}{1-x}$$
You can find $y=f'$ and hence $f$ from here. Plug in $x=\frac{1}{2}$.
A: This sum is in fact the series representation of $\operatorname{Li}_2\left(\frac12\right)$, which has a known closed form.
A: $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$
 Integrating from 0 to t we get
$$\int_{0}^{t}\frac{1}{(1-x)}dx=\sum_{n=0}^{\infty}\int_{0}^{t} x^{n}dx$$$$-\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n}$$
Dividing by t and integrating
$$\int_{0}^{0.5}-\frac{\ln(1-t)}{t}dt=\sum_{n=1}^{\infty}\int_{0}^{0.5} \frac{t^{n-1}}{n}dt=\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$
This on calculating is $$ \dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$$
A: I think a better way to go about this is as follows. Put $g(x)=-\int_0^x\frac{\ln(1-t)}t\text dt$. As you know from the other answers, you are trying to compute $\frac 12g(1/2)$. Now prove that $g(x)+g(1-x)=g(1)-\ln(x)\ln(1-x)$ using the definition of $g$, a substitution and integration by parts. Then expand the Taylor series for the integrand of $g(1)$ and integrate term by term to get $g(1)=\sum_{n\ge1}n^{-2}=\frac{\pi^2}6$ (this is called the Basel Problem). Finally, plug in $x=1/2$ and simplify.
