How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way? I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42 
\end{equation*}
using the identity 
$\displaystyle\frac{1}{1-x^2}\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)x^{2n-1}$
but kind of lengthy. any other approaches?
 A: Outstanding solution due to Cornel Valean. 
Recall the generating function $\displaystyle-\ln(1+x)\ln(1-x)=\sum_{n=1}^\infty x^{2n}\frac{H_{2n}-H_n}{n}+\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2},$ where if we multiply both sides by $\ln(1+x)/x$ and use the simple fact $\displaystyle\int_0^1x^{2n-1}\ln(1+x)\ dx=\frac{H_{2n}-H_n}{2n}$ then the Au-Yeung eries result, $\displaystyle\sum_{n=1}^\infty\left(\frac{H_n}{n}\right)^2=\frac{17}{4}\zeta(4),$ and $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$, we have
$$\small{\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}-2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}-2\sum_{n=1}^\infty\frac{H_n}{(2n)^3}=\frac{29}{16}\zeta(4)-\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx=\frac{23}{16}\zeta(4)}\tag{1}$$
where $\displaystyle\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx=-\frac38\zeta(4)$ is an already famous integral elementary to evaluate using the algebraic identity, $\displaystyle6a^2b=(a+b)^3-(a-b)^3-2b^3$. since $\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}{n^3}=\frac{11}4\zeta(4)-\frac74\ln(2)\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42-2\operatorname{Li}_4\left(\frac12\right)$ and $\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{n^2}=\frac{41}{16}\zeta(4)-\frac74\ln(2)\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42-2\operatorname{Li}_4\left(\frac12\right)$, if using for the last two series in (1) that $2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty (-1)^{n-1}a_n$, we conclude that $\displaystyle\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}=\frac{13}8\zeta(4)+\frac72\ln(2)\zeta(3)-\ln^22\zeta(2)+\frac1{6}\ln^42+4\operatorname{Li}_4\left(\frac12\right)$ and the solution is complete.
