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.
 A: 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}
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{%
\sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}}}
=\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n}
+\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}}
={\rm Li}_{1}\pars{\half} + {\rm Li}_{2}\pars{\half}
\end{align}
  where
  $\ds{{\rm Li_{s}}\pars{z} \equiv \sum_{k = 1}^{\infty}{z^{k} \over k^{\rm s}}}$
  is the
  PolyLogarithm Function.

However:
$$
{\rm Li}_{1}\pars{\half} = \ln\pars{2}\qquad\mbox{and}\qquad
{\rm Li}_{2}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}
$$

$$\color{#00f}{\large%
\sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}}
={\pi^{2} \over 12} + \ln\pars{2} - \half\,\ln^{2}\pars{2}}\approx 1.2754
$$

A: Consider Maclaurin series of natural logarithm
$$
\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}.
$$
Taking $x=\dfrac12$ yields
\begin{align}
\ln\left(1-\frac12\right)&=-\sum_{n=1}^\infty\frac{1}{2^n\ n}\\
\ln2&=\sum_{n=1}^\infty\frac{1}{2^n\ n}.
\end{align}
Now, dividing the Maclaurin series of natural logarithm by $x$ yields
\begin{align}
\frac{\ln(1-x)}{x}&=-\sum_{n=1}^\infty\frac{x^{n-1}}{n},
\end{align}
then integrating both sides and taking the limit of integration $0<x<\dfrac12$. We obtain
\begin{align}
\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=-\int_0^{\Large\frac12}\sum_{n=1}^\infty\frac{x^{n-1}}{n}\ dx\\
&=-\sum_{n=1}^\infty\int_0^{\Large\frac12}\frac{x^{n-1}}{n}\ dx\\
&=-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=0}^{\Large\frac12}\\
-\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=\sum_{n=1}^\infty\frac{1}{2^n\ n^2}.
\end{align}
The integral in the LHS is $\text{Li}_2\left(\dfrac12\right)=\dfrac{\pi^2}{12}-\dfrac{\ln^22}{2}$, but since you are not familiar with dilogarithm function, we will evaluate the LHS integral using IBP. Taking $u=\ln(1-x)$ and $dv=\dfrac1x\ dx$, we obtain
\begin{align}
\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=\left.\ln(1-x)\ln x\right|_0^{\large\frac12}+\int_0^{\Large\frac12}\frac{\ln x}{1-x}\ dx\\
&=\ln^22-\lim_{x\to0}\ln(1-x)\ln x-\int_1^{\Large\frac12}\frac{\ln (1-x)}{x}\ dx\ ;\\&\color{red}{\Rightarrow\text{let}\quad x=1-x}\\
\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx+\int_1^{\Large\frac12}\frac{\ln (1-x)}{x}\ dx&=\ln^22-0\\
-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=0}^{\Large\frac12}-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=1}^{\Large\frac12}&=\ln^22\\
\sum_{n=1}^\infty\frac{1}{2^n\ n^2}+\sum_{n=1}^\infty\frac{1}{2^n\ n^2}-\sum_{n=1}^\infty\frac{1}{n^2}&=-\ln^22\\
2\sum_{n=1}^\infty\frac{1}{2^n\ n^2}-\frac{\pi^2}{6}&=-\ln^22\\
\sum_{n=1}^\infty\frac{1}{2^n\ n^2}&=\frac{\pi^2}{12}-\frac{\ln^22}{2}.
\end{align}
Thus,
\begin{align}
\sum_{n=1}^\infty\frac{n+1}{2^n\ n^2}&=\sum_{n=1}^\infty\left(\frac{1}{2^n\ n}+\frac{1}{2^n\ n^2}\right)\\
&=\large\color{blue}{\ln2+\frac{\pi^2}{12}-\frac{\ln^22}{2}}.
\end{align}
