TITLE: Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$
QUESTION [32 upvotes]: How can I prove that 
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$
Can anyone help me please?

REPLY [4 votes]: \begin{align*}
\operatorname{Li}_3\left(\frac{1}{2}\right)&=\int _0^{\frac{1}{2}}\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx=-\operatorname{Li}_2\left(\frac{1}{2}\right)\ln \left(2\right)+\int _{\frac{1}{2}}^{1}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\int _0^1\frac{\ln \left(1-x\right)\ln \left(x\right)}{x}\:dx-\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\sum _{k=1}^{\infty }\frac{1}{k^3}-\frac{1}{2}\ln ^3\left(2\right)-\frac{1}{2}\int _0^{\frac{1}{2}}\frac{\ln ^2\left(1-x\right)}{x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\operatorname{Li}_3\left(\frac{1}{2}\right)
\end{align*}
And magically we find the value for that sum
\begin{align*}
\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)
\end{align*}