Sum $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}H_{2k}}{k}$ I am particularly interested in solving 
$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}H_{2k}}{k}$$
Where $$H_n = \sum_{k=1}^n \frac{1}{k}$$
I can’t seem to crack it. As much as I’d love to use the generating functions out there for harmonic number sums I can’t find one that suits this problem well - if the matter of solving it even is having a decent generating function.
 A: The generating function of the harmonic numbers is
$$
\sum_{k=1}^\infty H_kz^k=\frac{\log(1-z)}{z-1}\;.
$$
Thus
\begin{eqnarray}
\sum_{k=1}^\infty\frac{H_k}kz^k
&=&
\int_0^1\frac1z\sum_{k=1}^\infty H_kz^k\mathrm dz
\\
&=&
\int_0^1\frac{\log(1-z)}{z(z-1)}\mathrm dz
\\
&=&
\operatorname{Li}_2(x)+\frac12\log^2(1-x)\;.
\end{eqnarray}
Then
\begin{eqnarray}
\sum_{k=1}^\infty\frac{(-1)^{k-1}H_{2k}}k
&=&
-\left(\sum_{k=1}^\infty\frac{H_k}k\mathrm i^k+\sum_{k=1}^\infty\frac{H_k}k(-\mathrm i)^k\right)
\\
&=&
-\left(\operatorname{Li}_2(\mathrm i)+\frac12\log^2(1-\mathrm i)+\operatorname{Li}_2(-\mathrm i)+\frac12\log^2(1+\mathrm i)\right)
\\
&=&
-2\mathfrak R\left(\operatorname{Li}_2(\mathrm i)+\frac12\log^2(1+\mathrm i)\right)
\\
&=&
-2\left(-\frac{\pi^2}{48}+\frac12\left(-\frac{\pi^2}{16}+\frac{\log^22}4\right)\right)
\\
&=&\frac{5\pi^2}{48}-\frac{\log^22}4\;.
\end{eqnarray}
A: By integration by parts we have
$$H_n=\int_0^1\frac{1-x^n}{1-x}dx=\underbrace{-\ln(1-x)(1-x^n)|_0^1}_{0}-n\int_0^1 x^{n-1}\ln(1-x)dx$$
replace $n$ by $2n$ then multiply both sides by $\frac{(-1)^{n-1}}{n}$ and consider the summation from $n=1$ to $\infty$ we have
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{2n}}{n}=2\int_0^1 \frac{\ln(1-x)}{x}\sum_{n=1}^\infty(-x^2)^ndx=-2\int_0^1\frac{x\ln(1-x)}{1+x^2}dx$$
$$=2\Re\int_0^1\frac{i\ln(1-x)}{1-ix}dx\overset{1-x=y}{=}2\Re\int_0^1\frac{i\ln(y)}{1-i+iy}dx=2\Re\text{Li}_2\left(\frac{i}{i-1}\right)$$
By Landen's identity , we have
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{2n}}{n}=2\Re\bigg[-\text{Li}_2(i)-\frac12\ln^2(1-i)\bigg]=\frac58\zeta(2)-\frac14\ln^2(2)$$
where in the calculations we used the generalization
$$\int_0^1\frac{a\ln^n(x)}{1-a+ax}dx=(-1)^{n-1}n!\text{Li}_{n+1}\left(\frac{a}{a-1}\right)$$ which can be found in the book Almost (Impossible) Integrals, Sums and Series page 4.
Also note the generalization
$$\Re\text{Li}_r(i)=\Re\sum_{n=1}^\infty \frac{i^n}{n^r}=\sum_{n=1}^\infty\frac{(-1)^n}{(2n)^r}=-2^{-r} \eta(r)=-2^{-r}[(1-2^{1-r})\zeta(r)]=(2^{1-2r}-2^{-r})\zeta(r)$$
Which gives $$\Re\text{Li}_2(i)=-\frac18\zeta(2)$$
And for a complex number $z=x+iy=re^{i\theta}$, we have $\ln(z)=\ln(re^{i\theta})=\ln(r)+i\theta$
so the $\Re(z)=\ln(r)$ and $\Im(z)=\theta$ where $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}(y/x)$. In our case, we have $\ln(r)=\ln(\sqrt{1+1})=\frac12\ln(2)=\Re\ln(1-i)$ and $\theta=\tan^{-1}(-1)=-\frac{\pi}{4}=\Im \ln(1-i)$. Thus,
$$\ln(1-i)=\frac12\ln(2)-\frac{\pi}{4}i\Longrightarrow \Re\ln^2(1-i)=\frac14\ln^2(2)-\frac{\pi^2}{16}=\frac14\ln^2(2)-\frac38\zeta(2).$$

A different solution without using the complex number:
Set $x=1$ in the series expansion of $\arctan(x),$
$$\sum_{n=1}^\infty(-1)^n\frac{2H_{2n}-H_n}{n}x^{2n}=-2\arctan^2(x),$$
we get
$$\sum_{n=1}^\infty(-1)^n\frac{2H_{2n}-H_n}{n}=-2\arctan^2(1)=-2\left(\frac{\pi}{4}\right)^2=-\frac34\zeta(2)$$
substitute $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}=\frac12\ln^2(2)-\frac12\zeta(2)$, which follows from using the generating function:
$$\sum_{n=1}^\infty\frac{H_n}{n}x^n=\text{Li}_2(x)+\frac12\ln^2(1-x),$$
we get
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n}=\frac14\ln^2(2)-\frac38\zeta(2).$$
