$$\displaystyle \sum_{m=0}^{\infty}{\frac{{\left({H}_{m}^{(1)}\right)}^{2} - {H}_{m}^{(2)}}{{(m+1)}^{6}}}$$
where $ \displaystyle {{H}_{k}}^{(r)} = \sum_{i=1}^{k}{\frac{1}{{i}^{r}}} $
I have no idea how to approach to this problem. I want a full closed form and an idea about how to approach these types of problems(use contour integration? I don't know, I haven't tried).
Edit: I would like to have the answer in terms of zeta function.
Recalling the definition of Euler's sums $$s_{h}\left(m,n\right)=\sum_{k\geq1}\frac{H_{k}^{m}}{\left(k+1\right)^{n}} $$ $$\sigma_{h}\left(m,n\right)=\sum_{k\geq1}\frac{H_{k}^{\left(m\right)}}{\left(k+1\right)^{n}} $$ and using the fact that $H_{0}=0 $ $$\sum_{m\geq0}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}=\sum_{m\geq1}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}=s_{h}\left(2,6\right)-\sigma_{h}\left(2,6\right) $$ and they can be write in terms of not so nice combinations of zeta function (see for example here and here). Using the identity $22$ here, we have explicitly $$\sum_{m\geq0}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}=s_{h}\left(2,6\right)-\sigma_{h}\left(2,6\right)=14\zeta\left(8\right)+\zeta\left(2\right)\zeta\left(6\right)-3\sum_{k=0}^{4}\zeta\left(6-k\right)\zeta\left(k+2\right)+ $$ $$+\frac{1}{3}\sum_{k=2}^{4}\zeta\left(6-k\right)\sum_{j=1}^{k-1}\zeta\left(j+1\right)\zeta\left(k+1-j\right). $$