computing the series $\sum_{n=1}^\infty \frac{1}{n^2 2^n}$ $$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$
I am new in series thus I tried a pair of methods to compute but I couldn't
 A: The idea is to consider the power serie 
$$f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2} $$
The value of your serie is then $f(\frac{1}{2} )$
Now, how to find $f$? Derivate !
$$f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n} $$
Now you multiply by $x$ to find an usual power serie :
$$xf'(x) = \sum_{n=1}^\infty \frac{x^n}{n} = -\ln(1-x)$$
Hence
$$f(x) = \int \frac{\ln(1-x)}{x} dx$$
$$f(1/2) - f(1) = \int_{\frac{1}{2}}^1  \frac{\ln(1-x)}{x} dx = [\ln(1-x)\ln(x) ]_{1/2}^1 + \int_{\frac{1}{2}}^1  \frac{\ln(x)}{1-x} dx $$
$$= -\ln(1/2)^2 - \int_0^{\frac{1}{2}}  \frac{\ln(1-x)}{x} dx $$
$$= -\ln(1/2)^2 - ( f(1/2)-f(0) ) $$
But $f(1) = \frac{\pi^2}{6}$, so 
$$2 f(1/2) = \frac{\pi^2}{6} -\ln(1/2)^2 $$
And you have the result
A: 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!
A: I think you should integrate $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ a couple times. We see $$-\log(1-x) =\sum^\infty_{n=0} \frac{x^{n+1}}{n+1}.$$ Then dividing by $x$, we see $$-\frac{\log(1-x)}{x} = \sum^\infty_{n=0} \frac{x^n}{n+1}.$$ Then integrating from $0$ to $1/2$ gives $$\sum^\infty_{n=0} \frac{1}{(n+1)^22^{n+1}} = - \int^{1/2}_{0} \frac{\log(1-x)}{x} dx = \frac{1}{12}(\pi^2 - 6\log(2)^2).$$ I'm not exactly sure how to perform that integral; I used WolframAlpha.
