I managed here to prove that
$$\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}\\=672\zeta(9)-240\zeta(2)\zeta(7)-105\zeta(3)\zeta(6)-168\zeta(4)\zeta(5)+24\zeta^3(3)$$
Where $H_n^{(r)}=\sum_{k=1}^n\frac1{k^r}$ is the harmonic number and $\zeta$ is The Riemann zeta function.
Also I managed to find the last sum $$\sum_{n=1}^\infty \frac{H_n^{(4)}}{n^5}=5\zeta(4)\zeta(5)+35\zeta(2)\zeta(7)-\frac{125}2\zeta(9).$$ My question here, is it possible to find the other four sums? and if so, are they known in the literature?
