[Largesum_{n=1}^{infty}frac{left(H_{n}^{(2)}
ight)^{2}}{n^{2}}=frac{19}{24}zeta(6)+zeta^{2}(3)
]
(Largemathbf{Proof:})
We use the Abel's rearrangement over the (N)-th partial sum of the series,
[egin{align*}sumlimits_{n=1}^{N}frac{left(H_n^{(2)}
ight)^2}{n^2} &= sumlimits_{n=1}^{N-1} left[left(H_n^{(2)}
ight)^2-left(H_{n+1}^{(2)}
ight)^2
ight]sumlimits_{k=1}^{n}frac{1}{k^2}+left(H_N^{(2)}
ight)^2sumlimits_{k=1}^{N} frac{1}{k^2}\&= left(H_N^{(2)}
ight)^3 - sumlimits_{n=0}^{N-1} frac{left(H_n^{(2)}+H_{n+1}^{(2)}
ight)H_n^{(2)}}{(n+1)^2}\
&= left(H_N^{(2)}
ight)^3 - sumlimits_{n=1}^{N} frac{left(2H_n^{(2)}-dfrac{1}{n^2}
ight)left(H_n^{(2)}-dfrac{1}{n^2}
ight)}{n^2}\&= left(H_N^{(2)}
ight)^3 - sumlimits_{n=1}^{N} frac{1}{n^2}left(2left(H_n^{(2)}
ight)^2-3frac{H_n^{(2)}}{n^2}+frac{1}{n^4}
ight)\
&= frac{1}{3}left(H_N^{(2)}
ight)^3+sumlimits_{n=1}^{N}frac{H_n^{(2)}}{n^4}-frac{1}{3}sumlimits_{n=1}^{N}frac{1}{n^6}end{align*}]
I.e.,(displaystyle sumlimits_{n=1}^{infty}frac{left(H_n^{(2)}
ight)^2}{n^2} = frac{1}{3}zeta(2)^3+sumlimits_{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 sumlimits_{n=1}^{infty} frac{H_n^{(2)}}{n^4} = zeta(3)^2 - frac{1}{3}zeta(6)). Combining the results lead to,
[Largeoxed{displaystyle sumlimits_{n=1}^{infty} frac{left(H_n^{(2)}
ight)^2}{n^2} = color{blue}{zeta(3)^2 + frac{19}{24}zeta(6)}}
]