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.

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}
