How to get a closed form for the sum $\sum_{n=1}^{\infty}{\frac{H_{2n}}{2n^2}}$?

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Is there any technique to get a closed form for $\sum_{n=1}^{\infty}{\frac{H_{2n}}{2n^2}}$?

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An answer to this was contained in Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$, its just $\frac{11}{8}\zeta(3)$.