Harmonic Sum

Wonderful problem based on generating function . Here is my approch.

The generating function of Harmonic number is $\sum_{n=1}^{\infty} H_{n} x^n=\frac{\ln(1-x)}{x-1}\Rightarrow \sum_{n=1}^{\infty}H_{n} x^{n-1} =\frac{\ln(1-x)}{x(x-1)}$ and hence on integrating from $0$ to $x=-1$ we yield $\sum_{n=1}^{\infty}\frac{H_n}{n} (-1)^{n}=\int_0^{-1} \frac{\ln(1-x)}{x(x-1)}=\operatorname{Li}_2(-1)+\frac{\ln^2(2)}{2}\\=-\eta(2) +\frac{\ln^2(2)}{2}=-\left(1-\frac{1}{2}\right)\zeta(2)+\frac{\ln^2(2)}{2}=-\frac{\pi^2}{12}+\frac{\ln^2(2)}{2}$ multiply by $-1$ we have as desired series on left giving us $\displaystyle \frac{\pi^2}{12}-\frac{\ln^2(2)}{2}$ . Or following the above work wr can deduce the generating function as $\sum_{n=1}^{\infty}\frac{H_n}{n} x^{n}=\operatorname{Li}_2(x)+\frac{\ln^2(1-x)}{2}\cdots(1)$ We evaluate $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n} }{n}$ by setting $x=i$ and multiplying thoroughly by $-2$ in main generating function $(1)$ .That is $\Re\left(\sum_{n=1}^{\infty}\frac{-2H_n}{n} i^n\right)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n}}{n} = \Re(-2\operatorname{Li}_2(i)-\ln^2(1-i))$ Note that $\operatorname{Li}_2(i)=-\frac{1}{4}\eta(2)+i\beta(2)=-\frac{\pi^2}{48}+iG$ also $\ln^2(1-i)=\ln(\sqrt{2}e^{-\frac{i\pi}{4}})^2=\left(\frac{\ln 2}{2}-\frac{i\pi}{4}\right)^2=\frac{\ln^2 2}{4}-\frac{\pi^2}{16}-i\frac{\ln 2\pi}{4}$ and the eqauting with real part we have $\Re(-2\operatorname{Li}_2(i)-\ln^2(1-i))=\frac{\pi^2}{24}-\frac{\ln^22}{4}+\frac{\pi^2}{16}=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ and our desired answer is $\frac{5\pi^2}{48}-\frac{\ln^22}{4}-\frac{\pi^2}{12}+\frac{\ln^22}{2}=\frac{\pi^2}{48}+\frac{\ln^22}{4}$

Notation $\eta(.)$ is Dirichlet eta function and $\beta(.)$ is Dirichlet beta function.