Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ 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.
 A: 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 $x=-1$ :
$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
A: Different approach:
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\sum_{n=1}^\infty\frac{H_{2n}+\frac{1}{2n+1}}{n^2}$$
$$=\sum_{n=1}^\infty\frac{H_{2n}}{n^2}+\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}$$
where
$$\sum_{n=1}^\infty\frac{H_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}$$
$$=4\sum_{n=1}^\infty\frac{1}{2n}\left(-\int_0^1x^{2n-1}\ln(1-x)dx\right)$$
$$=-2\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{x^{2n}}{n}$$
$$=2\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}dx$$
$$=2\int_0^1\frac{\ln^2(1-x)}{x}dx+2\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$
$$=2(2\zeta(3))+2(-\frac58\zeta(3))=\frac{11}{4}\zeta(3)$$
and
$$\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{2n}dx$$
$$=\int_0^1\sum_{n=1}^\infty \frac{x^{2n}}{n^2}=\int_0^1\text{Li}_2(x^2)dx$$
$$=x\text{Li}_2(x^2)|_0^1+2\int_0^1\ln(1-x^2)dx$$
$$=\zeta(2)+2(x-1)\ln(1-x^2)|_0^1-4\int_0^1\frac{x}{1+x}dx$$
$$=\zeta(2)-4(1-\ln(2))$$
