$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$ by integration or differentiation There is an infinite sum given:
$$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$$
It should be solved using integration, derivation or both.
I think using power series can help but I don't know how to finish the calculation. Any help will be appreciated!
 A: My answer to a duplicate question says:
$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$
 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: Note that we have $\int_0^x t^{n-1}\,dt=\frac{x^n}{n}$.  Then, we can write
$$\begin{align}
\sum_{n=1}^\infty \frac{x^{2n}}{n^2}&=\sum_{n=1}^\infty \int_0^x t^{n-1}\,dt\int_0^x s^{n-1}\,ds\\\\
&=\int_0^x\int_0^x \frac{1}{1-st}\,ds\,dt\\\\
&=-\int_0^{x} \frac{\log(1-sx)}{s}\,ds\\\\
&=-\int_0^{x^2} \frac{\log(1-s)}{s}\,ds\\\\
&=\text{Li}_2(x^2)
\end{align}$$
Evaluating at $x=1/\sqrt{2}$ yields
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^2\,2^n}=\text{Li}_2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)}$$
And we are done!

To evaluate $\text{Li}_2(1/2)$, we exploit the relationship
$$\text{Li}_2(1-x)=-\text{Li}_2\left(1-\frac1x\right)-\frac12\log^2(x)$$
Letting $x=1/2$ yields
$$\text{Li}_2(1/2)=-\text{Li}_2\left(-1\right)-\frac12\log^2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)$$
where we used 
$$\begin{align}
\text{Li}_2(-1)&=-\int_0^{-1}\frac{\log(1-x)}{x}\,dx\\\\
&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\\\\
&=\frac{\pi^2}{12}
\end{align}$$
A: Another way to calculate it: consider
$$
f(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n^2}.
$$
Differentiating w.r.t. $x$ gives
$$
f'(x)=\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n}=\frac{1}{x}\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad xf'(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad
(xf'(x))'=\sum_{n=1}^{+\infty}x^{n-1}=\frac{1}{1-x}.
$$
Now integrating with $f(0)=0$
$$
xf'(x)=-\ln(1-x)\quad\Rightarrow\quad f(x)=-\int_0^x\frac{\ln(1-t)}{t}\,dt.
$$
Motivation for termwise differentiation for power series is straightforward.
