Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ Finding the closed form of:
$$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$
where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$
It appears when we try to determine the summation $\displaystyle \sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, using the generating function:
\begin{align}
\sum_{n=1}^{\infty} \frac{H_n}{n^3} \, x^{n} &= - \frac{1}{2} \, \sum_{n=1}^{\infty}\frac{1}{n^2} \, \sum_{k=1}^{n} \frac{(1-x)^k}{k^2} - \frac{\zeta(2)}{2} \, \operatorname{Li}_2(x) + \frac{7 \, \zeta(4)}{8} - \frac{1}{4} \, \operatorname{Li}_2^2(1-x) + \frac{\zeta^2(2)}{4} + \operatorname{Li}_4(x) \\
& \hspace{5mm} + \frac{1}{4} \, \log^2 x \, \log^2(1-x) + \frac{1}{2}\log x \, \log (1-x) \, \operatorname{Li}_2(1-x) + \zeta(3) \, \log x - \log x \, 
\operatorname{Li}_2(1-x)
\end{align}
when we write, $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$
Combined with Cleo's closed form here, I know what the closed form should be, but how do I derive the result ?
 A: Different approach:
By Cauchy product we have 
$$\operatorname{Li}^2_2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$
take $x=1/2$ and rearrange the terms to get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}=\frac12\operatorname{Li}^2_2\left(\frac12\right)+3\operatorname{Li}_4\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^32^n}$$
Plugging $
\sum_{n=1}^\infty \frac{H_n}{n^32^n}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$ and $\operatorname{Li_2}\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$
gives 

\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42
\end{align*}

A: HINT: Consider $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{2^nn^2}+\operatorname{Li}_4\left(\frac{1}{2}\right)$ and then express the remaining sum as a double integral. After some work, you get 
$$\int_0^1 \frac{\displaystyle\log(x)\operatorname{Li}_2\left(\frac{x}{2}\right)}{x-2} \ dx+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
and after letting $x\mapsto 2x$ combined with the integration by parts, you arrive at some integrals
pretty easy to finish. I'm confident you can finish the rest of the job to do. 
$$\frac{\pi^4}{1440}-\frac{\pi^2}{3}\log^2(2)+\frac{1}{24}\log^4(2)+\frac{7}{24}\pi^2\log^2(2)+\frac{1}{4}\log(2)\zeta(3)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
