sequences and series - Sum of Harmonic numbers $sumlimits_{n=1}^{infty} frac{H_n^{(2)}}{2^nn^2}$

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 ?

Answer

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.

$$\frac{\pi^4}{1440}-\frac{\pi^2}{3}\log^2(2)+\frac{1}{24}\log^4(2)+\frac{7}{24}\pi^2\log^2(2)+\frac{1}{4}\log(2)\zeta(3)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$