TITLE: Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $
QUESTION [10 upvotes]: Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$

I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another question when the $n^2$ is in the numerator, but no significant progress so far.
Any help will be appreciated.
Thanks!

REPLY [7 votes]: Recall that
$$
\sum_{k=1}^\infty\frac{x^k}k=-\log(1-x)\tag{1}
$$
Dividing $(1)$ by $x$ and integrating yields
$$
\begin{align}
\sum_{k=1}^\infty\frac1{2^kk^2}
&=-\int_0^{1/2}\frac{\log(1-x)}{x}\,\mathrm{d}x\tag{2a}\\
&=-\left[\vphantom{\int}\log(1-x)\log(x)\right]_0^{1/2}
-\int_0^{1/2}\frac{\log(x)}{1-x}\,\mathrm{d}x\tag{2b}\\
&=-\log(2)^2-\int_{1/2}^1\frac{\log(1-x)}x\,\mathrm{d}x\tag{2c}\\
&=-\frac{\log(2)^2}2-\frac12\int_0^1\frac{\log(1-x)}x\,\mathrm{d}x\tag{2d}\\
&=-\frac{\log(2)^2}2+\frac12\sum_{k=1}^\infty\frac1{k^2}\tag{2e}\\
&=\bbox[5px,border:2px solid #00A000]{-\frac{\log(2)^2}2+\frac{\pi^2}{12}}\tag{2f}
\end{align}
$$
Explanation:
$\text{(2a)}$: divide $(1)$ by $x$ and integrate over $[0,1/2]$
$\text{(2b)}$: integrate by parts
$\text{(2c)}$: substitute $x\mapsto1-x$
$\text{(2d)}$: average $\text{(2a)}$ and $\text{(2c)}$
$\text{(2e)}$: divide $(1)$ by $x$ and integrate over $[0,1]$
$\text{(2f)}$: $\zeta(2)=\frac{\pi^2}6$