How to prove only with real methods that $$\displaystyle \sum _{n=1}^{\infty }\frac{\left(-1\right)^nH_{2n}}{n^2}=\frac{23}{16}\zeta (3)-\pi G$$ This sum can be evaluated by doing the following $$\sum _{n=1}^{\infty }\frac{\left(-1\right)^nH_{2n}}{n^2}=4\operatorname*{\mathfrak{R}} \sum _{n=1}^{\infty }\frac{i^nH_n}{n^2}$$ And then using the generating function $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ Setting $x=i$ and taking the real part of each term, but how can i evaluate this without resorting to complex methods?.
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The series is equal to $\displaystyle -\int_0^1 \frac{\frac{\pi^2}{12}+\text{Li}_2\left(-(1-x)^2\right)}{x}dx$ – FDP Sep 21 '20 at 09:12
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@FDP may I ask how you did get this integral from the original series? I have a really similar integral and was hoping to convert it into a series with harmonic numbers that maybe can be evaluated. Thank you. – Zima Jul 07 '23 at 15:41
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@Zima: \begin{align}S&=\sum {n=1}^{\infty }\frac{\left(-1\right)^nH{2n}}{n^2}\ &=\sum {n=1}^{\infty }\frac{(-1)^n}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}dx\right)\ &\int_0^1 \frac{1}{1-x}\left(\sum{n=1}^{\infty}\frac{(-1)^n}{n^2}-\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n}}{n^2}\right)dx\ &=-\int_0^1 \frac{1}{1-x}\left(\frac{\pi^2}{12}+\text{Li}_2\left(-x^2\right)\right)dx\overset{u=1-x}=-\int_0^1 \frac{\frac{\pi^2}{12}+\text{Li}_2\Big(-(1-u^2)\Big)}{u}du \end{align} – FDP Jul 14 '23 at 20:52