TITLE: Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$
QUESTION [36 upvotes]: 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? 
$$\sum_{n=1}^{\infty}  \frac{H_{
n}}{2^nn^4}$$

REPLY [2 votes]: Another solution using nice integral manipulations
From this solution we have that 
$$\small{I=\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx=\frac3{16}\zeta(5)+\frac3{20}\ln^52-\frac14\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx+\frac12\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx}$$
where
\begin{align}
\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx&=\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\
&=\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\
&=\small{24\zeta(5)-24\operatorname{Li}_5\left(\frac12\right)-24\ln2\operatorname{Li}_4\left(\frac12\right)-12\ln^22\operatorname{Li}_3\left(\frac12\right)-4\ln^32\operatorname{Li}_2\left(\frac12\right)-\ln^52}\\
\end{align}
and 
\begin{align}
\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx&=\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^3x\ dx\\
&=6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=6\sum_{n=1}^\infty\frac{H_n}{n^4}-6\zeta(5)
\end{align}
combine the two integrals  
$$\small{I=\frac25\ln^52-\frac{141}{16}\zeta(5)+6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)+3\ln^22\operatorname{Li}_3\left(\frac12\right)+\ln32\operatorname{Li}_2\left(\frac12\right)+3\sum_{n=1}^\infty\frac{H_n}{n^4}}\tag{1}$$

On the other hand
\begin{align}
I&=\int_{1/2}^{1}\frac{\ln^3(1-x)\ln x}{x}\ dx\overset{x\mapsto 1-x}{=}\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{1-x}\ dx\\
&=\sum_{n=1}^\infty H_n\int_0^{1/2}- x^{n}\ln^3x\ dx=\sum_{n=1}^\infty \left(H_n-\frac1n\right)\int_0^{1/2} -x^{n-1}\ln^3x\ dx\\
&=\sum_{n=1}^\infty \left(H_n-\frac1n\right)\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}\right)\tag{2}
\end{align}

From (1) and (2) we have that 
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n}{2^nn^4}
&=-\ln2\sum_{n=1}^{\infty}\frac{H_n}{2^n n^3}-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{2^n n^2}-\frac16\ln^32\sum_{n=1}^{\infty}\frac{H_n}{2^n n}+\frac12\sum_{n=1}^{\infty}\frac{H_n}{n^4}-\frac{47}{32}\zeta(5)\\
&\quad+\frac{1}{15}\ln^52+\frac{1}{3}\ln^32\operatorname{Li_2}\left( \frac12\right)+\ln^22\operatorname{Li_3}\left( \frac12\right)+2\ln2\operatorname{Li_4}\left( \frac12\right) +2\operatorname{Li_5}\left( \frac12\right)
\end{align*}
Substituting   
$$
S_1=\sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$$
$$S_2=\sum_{n=1}^{\infty}\frac{H_n}{2^n n^2}=\zeta(3)-\frac{1}{2}\ln(2)\zeta(2)$$
$$S_3=\sum_{n=1}^{\infty}\frac{H_n}{2^n n}=\frac12\zeta(2)$$
along with $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3),\ $ $\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$ and $\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$ 
gives

\begin{align}
\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{2^n n^4}&=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: $S_1$ can be found here and $S_2$ and $S_3$ can be found here.