TITLE: Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
QUESTION [19 upvotes]: How can I prove that 
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.

REPLY [2 votes]: Using the following nice rule: $$\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}\left(\frac{1+(-1)^n}{2}\right)$$
We get
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n+1}}{{(2n)}^2}=4\left(\frac12\sum_{n=1}^\infty\frac{H_{n+1}}{n^2}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n+1}}{n^2}\right)\\
&=2\sum_{n=1}^\infty\frac{H_n}{n^2}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}+2\sum_{n=1}^\infty\frac{1}{n^2(n+1)}+2\sum_{n=1}^\infty\frac{(-1)^n}{n^2(n+1)}\\
&=2\left(2\zeta(3)\right)+2\left(-\frac58\zeta(3)\right)+2\left(\zeta(2)-1\right)+2\left(2\ln2-\frac12\zeta(2)-1\right)\\
&=\frac{11}4\zeta(3)+\zeta(2)+4\ln2-4
\end{align}
Note that $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}$ was obtained from using the generating function where we set $x=-1$ :
$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$