Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$ One of the possible ways of computing the series is to obtain the generating function, but
this might be a tedious, hard work, pretty hard to obtain. What would you propose then? 
$$\sum_{n=1}^{\infty}  \frac{H_{
n}}{2^nn^4}$$
 A: Here is a magical solution:
We proved here
\begin{align}
I&=\int_0^1\frac{\ln^2(1-x)}{1-x}\left(\ln^2(1+x)-\ln^2(2)\right)\ dx\\
&=\small{\boxed{\frac18\zeta(5)-\frac12\ln2\zeta(4)+2\ln^22\zeta(3)-\frac23\ln^32\zeta(2)-2\zeta(2)\zeta(3)+\frac1{10}\ln^52+4\operatorname{Li}_5\left(\frac12\right)\quad}}\tag{1}
\end{align}
On the other hand and by integration by parts, we have
\begin{align}
I&=\frac23\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{1+x}\ dx\overset{\color{red}{1-x\ \mapsto\ x}}{=}\frac13\int_0^1\frac{\ln^3x\ln(2-x)}{1-x/2}\ dx\\
&=\frac{\ln2}{3}\int_0^1\frac{\ln^3x}{1-x/2}\ dx+\frac13\int_0^1\frac{\ln^3x\ln(1-x/2)}{1-x/2}\ dx\\
&=\frac{\ln2}{3}\sum_{n=1}^\infty\frac{1}{2^{n-1}}\int_0^1x^{n-1}\ln^3x\ dx-\frac13\sum_{n=1}^\infty\frac{H_n}{2^n}\int_0^1x^n\ln^3x\ dx\\
&=\frac{\ln2}{3}\sum_{n=1}^\infty\frac{1}{2^{n-1}}\left(-\frac{6}{n^4}\right)-\frac13\sum_{n=1}^\infty\frac{H_n}{2^n}\left(-\frac{6}{(n+1)^4}\right)\\
&=-4\ln2\sum_{n=1}^\infty\frac{1}{n^42^n}+2\sum_{n=1}^\infty\frac{H_n}{(n+1)^42^n}\\
&=\boxed{-4\ln2\operatorname{Li}_4\left(\frac12\right)+4\sum_{n=1}^\infty\frac{H_n}{n^42^n}-4\operatorname{Li}_5\left(\frac12\right)}\tag{2}
\end{align}
From $(1)$ and $(2)$, we get
\begin{align}
\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2)
+\frac12\ln^22\zeta(3)\\
&\quad-\frac18\ln2\zeta(4)- 
\frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52
\end{align}

Note: Full credit goes to Cornel for proposing such amazing problem in $(1)$.
