Closed form of $\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$ Find the closed form of $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$$
I know the closed form for smaller powers like $2, 3$ exists, but I'm not sure if there is a
closed form  for this variant. Is it possible to tackle the question in an elementary way
and find the answer, without using integrals at all?
Then, if this exists, I'd also propose the alternating variant
$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\frac{H_n}{n}\right)^4$$

Update (by editor): The last sum by OP is:
$$\small \sum _{n=1}^{\infty } (-1)^{n-1} \left(\frac{H_n}{n}\right){}^4=\frac{633}{128} \zeta(6,2)-\frac{1}{6} \pi ^2\zeta(\bar5,1)+6 \zeta(\bar7,1)-4 \zeta(\bar5,1,\bar1,1)+6 \log ^2(2) \zeta(\bar5,1)+12 \log (2) \zeta(\bar5,1,1)-16 \text{Li}_5\left(\frac{1}{2}\right) \zeta (3)-\frac{1}{9} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^2 \zeta (3)^2}{8}-\frac{63 \zeta (3) \zeta (5)}{64}+\frac{2}{15} \zeta (3) \log ^5(2)-\frac{2}{9} \pi ^2 \zeta (3) \log ^3(2)-\frac{31}{12} \zeta (5) \log ^3(2)+3 \zeta (3)^2 \log ^2(2)-\frac{16}{45} \pi ^4 \zeta (3) \log (2)-\frac{65}{48} \pi ^2 \zeta (5) \log (2)+\frac{155}{8} \zeta (7) \log (2)+\frac{149141 \pi ^8}{43545600}-\frac{1}{216} \pi ^4 \log ^4(2)+\frac{25 \pi ^6 \log ^2(2)}{1512}$$

The result is
\begin{align*}
S_{1^4,4}=\sum\limits_{n = 1}^\infty  {\frac{{H_n^4}}{{{n^4}}}}  = \frac{{13559}}{{144}}\zeta \left( 8 \right) - 92\zeta \left( 3 \right)\zeta \left( 5 \right) - 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 26{S_{2,6}},
\end{align*}
Detailed process see the paper ``Multiple zeta values and Euler sum"  enter link description here
