I want to find the sum $$\sum_{k=1}^{\infty}\frac{H^2_n}{n^2}$$
For some background context. I recently found the Taylor series of Dilogarithm squared as
$$\mathrm{Li}_2(x)^2=\sum_{n=2}^{\infty}\left(\frac{2nH^{(2)}_{n-1}+4H_{n-1}}{n^3}\right)x^n$$
Where $$H_n^{(m)}=\sum_{k=1}^{n}\frac{1}{k^m}$$
This can be derived from expanding $\mathrm{Li}_2(x)$ in series form, square it and use cauchy product
The interesting thing is that if $x=1$ I found that
$$\sum_{k=1}^{\infty}\frac{H^{(2)}_n}{n^2}=\frac{7\pi^4}{360}$$
So I'm wondering if I can find the sum
$$\sum_{k=1}^{\infty}\frac{H^2_n}{n^2}$$
I've tried to use the fact that
$$\frac{H^{(2)}_n+H^2_n}{n}=\frac{2}{n}\sum_{k=1}^{n}\frac{H_k}{k}$$
and take the sum on both sides but failed because I couldn't handle the sum right hand side.
Any help is appreciated
