Series Question: $\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$$
I tried
$$\frac{n+1}{2^nn^2}=\frac{1}{2^nn}+\frac{1}{2^nn^2}$$
The idea is using Riemann zeta function
$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
but the term $2^n$ makes complicated. I know that
$$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$
using geometric series but I don't know how to use those series to answer the question. Any help would be appreciated. Thanks in advance.

The series 
\begin{align}
S = \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} 
\end{align}
can be expressed as 
\begin{align}
S = \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n} + \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n^{2}} 
\end{align}
and is seen to be
\begin{align}
S = - \ln\left( 1 - \frac{1}{2} \right) + Li_{2}\left(\frac{1}{2}\right),
\end{align}
where $Li_{2}(x)$ is the dilogarithm function. Since
\begin{align}
Li_{2}\left(\frac{1}{2}\right) = \frac{\pi^{2}}{12} - \frac{1}{2} \ \ln^{2}(2)
\end{align}
then the resulting series has the value
\begin{align}
\sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \ln(2) - \frac{1}{2} \ \ln^{2}(2).
\end{align}
This may also be seen in the form
\begin{align}
\sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \frac{1}{2} \ \ln(2) \ \ln\left(\frac{e^{2}}{2}\right).
\end{align}
