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

Question

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

Answer 1

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