In one of the answers to this question Evaluate $\int_0^1\arcsin^2(\frac{\sqrt{-x}}{2}) (\log^3 x) (\frac{8}{1+x}+\frac{1}{x}) \, dx$, the OP also presents some results from a paper, many of them obviously related to sums of harmonic series of weight $6$ (perhaps also magical cancellations involving some of the closed forms prevent me from saying more).
Here is an example of a sum that is strongly related to those sums - in fact, to solve those questions elementarily one wants to get transformations (this may take some time and it can be a boring task) to such forms as the one below (this is just one example)
\begin{equation*} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n}{n^5} -\frac{3}{2}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^4} +\frac{4}{3}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^3}{n^3}-\frac{2}{3}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^4}{n^2} \end{equation*} \begin{equation*} +\frac{1}{6}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(3)}}{n^3}-\frac{1}{3}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n H_n^{(3)}}{n^2} \end{equation*} \begin{equation*} =\frac{19}{128} \zeta (6)+\frac{1}{16}\zeta^2(3). \end{equation*}
Since the previous question with such questions had been well received, maybe the community will enjoy some more sums of series alike, leading to very simple and neat closed forms.
The sum of the series above is proposed by C. I. Valean, and it is obtained very elegantly - such a solution will be presented here after the release of More (Almost) Impossible Integrals, Sums, and Series, the sequel of (Almost) Impossible Integrals, Sums, and Series.
How would you attack it?
