Euler squared sum of order 2 : $\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}$ Let $\mathcal{H}_n$ denote the $n$ - th harmonic number. What techniques would one use to prove that
$$\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2} = \zeta^2(3) + \frac{19 \zeta(6)}{24}$$
where $\zeta$ denotes the Riemann zeta function.  

Since 
$$\left(\mathcal{H}_n^{(2)}\right)^2=\mathcal{H}_n^{(4)}+2\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}$$
we can rewrite the sum in terms of Multiple zeta functions:
\begin{align*}\sum_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n^{(4)}}{n^2}+
2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}\\
&=\left(\zeta(6)+\zeta(2,4)\right)+2(\zeta(2,2,2)+\zeta(4,2))\\
&=\zeta(6)+\left(\frac{25}{12}\zeta(6)-\zeta(3)^2\right)+2\left(\frac{\zeta(2)^3}{6}+\frac{\zeta(6)}{3}-\frac{\zeta(2)\zeta(4)}{2}\right)\\
&\qquad+2\left(\zeta(3)^2-\frac{4\zeta(6)}{3}\right)\\
&=\zeta(3)^2+\frac{19\pi^6}{22680}=\zeta(3)^2 + \frac{19 \zeta(6)}{24}
\end{align*}
where we also used the explicit values of 
$$\zeta(2)=\frac{\pi^2}{6},\quad \zeta(4)=\frac{\pi^4}{90}, \quad \zeta(6)=\frac{\pi^6}{945}.$$
