\[\Large\sum_{n=1}^{\infty}\frac{\left(H_{n}^{(2)}\right)^{2}}{n^{2}}=\frac{19}{24}\zeta(6)+\zeta^{2}(3)\]
\(\Large\mathbf{Proof:}\)
We use the Abel's rearrangement over the \(N\)-th partial sum of the series,
\[\begin{align*}\sum\limits_{n=1}^{N}\frac{\left(H_n^{(2)}\right)^2}{n^2} &= \sum\limits_{n=1}^{N-1} \left[\left(H_n^{(2)}\right)^2-\left(H_{n+1}^{(2)}\right)^2\right]\sum\limits_{k=1}^{n}\frac{1}{k^2}+\left(H_N^{(2)}\right)^2\sum\limits_{k=1}^{N} \frac{1}{k^2}\\&= \left(H_N^{(2)}\right)^3 - \sum\limits_{n=0}^{N-1} \frac{\left(H_n^{(2)}+H_{n+1}^{(2)}\right)H_n^{(2)}}{(n+1)^2}\\ &= \left(H_N^{(2)}\right)^3 - \sum\limits_{n=1}^{N} \frac{\left(2H_n^{(2)}-\dfrac{1}{n^2}\right)\left(H_n^{(2)}-\dfrac{1}{n^2}\right)}{n^2}\\&= \left(H_N^{(2)}\right)^3 - \sum\limits_{n=1}^{N} \frac{1}{n^2}\left(2\left(H_n^{(2)}\right)^2-3\frac{H_n^{(2)}}{n^2}+\frac{1}{n^4}\right)\\ &= \frac{1}{3}\left(H_N^{(2)}\right)^3+\sum\limits_{n=1}^{N}\frac{H_n^{(2)}}{n^4}-\frac{1}{3}\sum\limits_{n=1}^{N}\frac{1}{n^6}\end{align*}\]
I.e.,\(\displaystyle \sum\limits_{n=1}^{\infty}\frac{\left(H_n^{(2)}\right)^2}{n^2} = \frac{1}{3}\zeta(2)^3+\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^4}-\frac{1}{3}\zeta(6)\)
M.N.S.E showed in this answer one way of dealing with \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^4} = \zeta(3)^2 - \frac{1}{3}\zeta(6)\). Combining the results lead to,
\[\Large\boxed{\displaystyle \sum\limits_{n=1}^{\infty} \frac{\left(H_n^{(2)}\right)^2}{n^2} = \color{blue}{\zeta(3)^2 + \frac{19}{24}\zeta(6)}}\]
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