Evaluating $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$ I would appreciate to understand the main steps giving the evaluation of this series: $$ S=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$$ where $H_n$ is the harmonic number. I've tried with no success to obtain this sum with the help of Wolfram Alpha.
 A: From the  generating function for the harmonic numbers, we immediately have that
$$\sum_{n=1}^{\infty}  \frac{H_n}{n}z^n=\frac{1}{2}\log^2(1-z)+\operatorname{Li_2}(z)$$
and after multiplying both sides by $-2$ and setting $z=i$, we get 
$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n}}{n}=-2\left(\frac{1}{2}\log^2(1-i)+\operatorname{Li_2}(i)\right)$$
$$=\frac{5\pi^2}{48}-\frac14 \ln^22$$
Q.E.D.
A: Using integral representation for harmonic number given by Leonhard Euler
\begin{equation}
H_k=\int_0^1\frac{1-x^k}{1-x}\,dx
\end{equation}
and Taylor series for natural logarithm
\begin{equation}
\ln(1+y)=\sum_{k=1}^\infty\, (-1)^{k+1}\frac{x^k}{k}\qquad,\qquad\mbox{for}\,\,|y|<1
\end{equation}
we have
\begin{align}
S&=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}\\
&=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_0^1\frac{1-x^{2n}}{1-x}\,dx\\
&=\int_0^1\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left[1-x^{2n}\right]\,\frac{dx}{1-x}\\
&=\int_0^1\left[\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\right]\,dx\\
\end{align}
Now, apply integration by parts by taking $u=\ln2-\ln\left(1+x^2\right)$ and $dv=\dfrac{dx}{1-x}$. We obtain
\begin{equation}
S=-\lim_{t\to1}\left.\left[\ln2-\ln\left(1+x^2\right)\right]\ln(1-x)\right|_0^t-2\int^1_0\frac{x\ln(1-x)}{1+x^2}\,dx
\end{equation}
Since we have an indeterminate form $0\cdot\infty$ as $t\to1$, we may write our limit as
\begin{equation}
-\lim_{t\to1}\frac{\ln(1-t)}{\frac{1}{\ln2-\ln\left(1+t^2\right)}}
\end{equation}
then apply L'Hospital's rule twice to obtain
\begin{equation}
\lim_{t\to1}\left[\ln2-\ln\left(1+t^2\right)\right]\ln(1-t)=0
\end{equation}
Therefore, we may write our original sum as
\begin{equation}
S=-2\int^1_0\frac{x\ln(1-x)}{1+x^2}\,dx
\end{equation}
The evaluation of the latter integral can be seen here. I take no credit for it. There are some good answers, but my personal choice is the answer using differentiation under the integral sign method since it's very simple and easy to understand. Anyway, this is a good question, thanks for sharing. ツ
A: Here is an approach to obtain

$$
\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}=\frac{5\pi^2}{48}-\frac14 \ln^22. \tag1
$$

Recall the standard series expansion
$$
\log(1+z)= \sum_{n=1}^{\infty} (-1)^{n-1}\frac{z^{n}}{n},\quad |z|\leq 1, z\neq -1. \quad (*)
$$
Then, by the Cauchy product, one has
$$
\begin{align}
\log(1+z)\log(1-z)
&=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{z^{n}}{n}\sum_{n=1}^{\infty} \frac{-z^{n}}{n}\\
&=\sum_{n=1}^{\infty}\left( \sum_{k=1}^{n-1} \frac{(-1)^{k}}{k(n-k)}\right)z^{n}\\
&=\sum_{n=1}^{\infty}\!\left( \frac{1+(-1)^{n}}{n}\sum_{k=1}^{n-1} \frac{(-1)^{k}}{k}\right)\!z^{n}\\
&=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{2n-1} \frac{(-1)^{k}}{k}z^{2n}\\
&=\sum_{n=1}^{\infty}\frac{1}{n}\!\left( H_{2n}-H_n-\frac1n\right)\!z^{2n}\\
\end{align}
$$
equivalently
$$
\begin{align}
\sum_{n=1}^{\infty}\frac{ H_{2n}}{n}z^{2n}=\log(1+z)\log(1-z)+\sum_{n=1}^{\infty}\frac{ H_{n}}{n}z^{2n}+\sum_{n=1}^{\infty}\frac{ 1}{n^2}z^{2n}
\end{align} \tag2
$$
Now, if you put $z:=i\,$ ($i^2=-1$) in $(2)$ you easily conclude with
$$
\begin{align}
& \log(1+i)=\log \left(\sqrt{2}\:e^{i\pi/4} \right)=\frac12 \ln2 + i\frac\pi 4 \\
& \log(1+i)\log(1-i)=\frac14 \ln^22+\frac{\pi^2}{16}\\
& \sum_{n=1}^{\infty}(-1)^{n}\frac{H_{n}}{n} =\frac{\ln^22}{2}-\frac{\pi^2}{12} \tag3\\
& \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2} =-\frac{\pi^2}{12} \tag4
\end{align}
$$
giving the desired result. 

We may prove $(3)$ by considering the standard identity, easily obtained from $(*)$,  $$\sum_{n=1}^{\infty}(-1)^{n}H_n x^{n-1} = -\dfrac{\ln(1+x)}{x(1+x)} \quad -1 < x<1,\,x\neq0,$$ integrating from $x=0^+$ to $x=1^-$ getting $$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} \!= \!- \int_{0}^{1}\dfrac{\ln(1+x)}{x(1+x)} dx = \!-\int_{0}^{1}\left(\dfrac{\ln(1+x)}{x}\! - \!\dfrac{\ln(1+x)}{1+x}\right) \! dx= -\dfrac{\pi^2}{12}+\dfrac{\ln^2 2}{2}$$ where we have classically used $(*)$ again to obtain $$\int_{0}^{1}\dfrac{\ln(1+x)}{x}dx = \dfrac{\pi^2}{12}. $$
