TITLE: computing the series $\sum_{n=1}^\infty \frac{1}{n^2 2^n}$
QUESTION [5 upvotes]: $$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$
I am new in series thus I tried a pair of methods to compute but I couldn't

REPLY [5 votes]: Note that we can write
$$\begin{align}
\sum_{n=1}^N\frac{x^n}{n^2}&=\sum_{n=1}^N \int_0^x s^{n-1}\,ds \int_0^1 t^{n-1}\,dt\\\\
&=\int_0^1 \int_0^x \frac{1-(st)^N}{1-(st)}\,ds\,dt \tag 1\\\\
\end{align}$$
For $x<1$, using the Dominated Convergence Theorem to evaluate the limit of $(1)$ as $N\to \infty$, we can write
$$\begin{align}
\sum_{n=1}^\infty \frac{x^n}{n^2}&=\int_0^1 \int_0^x \frac{1}{1-(st)}\,ds\,dt\\\\
&=-\int_0^1 \frac{\log(1-xt)}{t}\,dt\\\\
&=-\int_0^x \frac{\log(1-u)}{u}\,du\\\\
&=\text{Li}_2(x) \tag 2
\end{align}$$
where in $(2)$ $\text{Li}_2(x)$ is the dilogarithm function.  Therefore, for $x=1/2$ we have
$$\sum_{n=1}^\infty \frac{1}{n^22^n}=\text{Li}_2(1/2) \tag 3$$
Note that $\text{Li}_2(1)=\frac{\pi^2}{6}$ See Basel Problem.  Furthermore, it can be shown (see the NOTE at the end of this solution) that the dilogarithm function satisfies the reflection identity 
$$\begin{align}
\text{Li}_2(x)+\text{Li}_2(1-x)&=\text{Li}_2(1)-\log(x)\log(1-x)\\\\
&=\frac{\pi^2}{6}-\log(x)\log(1-x) \tag 4
\end{align}$$
Using $(4)$ with $x=1/2$ in $(3)$ reveals
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^22^n}=\frac{\pi^2}{12}-\frac12 \log^2(1/2)}$$
And we are done!

NOTE:
Here, we prove the reflection identity given in $(4)$.  Observe that we can write
$$\begin{align}
\text{Li}_2(x)+\text{Li}_2(1-x)&=-\int_0^x \frac{\log(1-u)}{u}\,du-\int_0^{1-x}\frac{\log(1-u)}{u}\,du\\\\
&=-\int_0^x \frac{\log(1-u)}{u}\,du-\int_0^{1}\frac{\log(1-u)}{u}\,du-\int_1^{1-x} \frac{\log(1-u)}{u}\,du\\\\
&=\frac{\pi^2}{6}-\int_0^x \frac{\log(1-u)}{u}\,du-\int_1^{1-x} \frac{\log(1-u)}{u}\,du \\\\
&=\frac{\pi^2}{6}-\int_0^x \frac{\log(1-u)}{u}\,du+\int_0^{x} \frac{\log(u)}{1-u}\,du \\\\
&=\frac{\pi^2}{6}-\int_0^x \frac{\log(1-u)}{u}\,du+\left.\left(-\log(u)\log(1-u)\right)\right|_0^x +\int_0^x \frac{\log(1-u)}{u}\,du\\\\
&=\frac{\pi^2}{6}-\log(x)\log(1-x) 
\end{align}$$
as was to be shown!