How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$, 
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\ 
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form. 
Can you suggest any ideas how to calculate $S$?
 A: let $$y=\sum_{n=1}^{\infty}H^2_{n}x^n$$
then we have 
$$y=x+xy+\ln^2{(1-x)}+\int_{0}^{x}\dfrac{\ln{(1-t)}}{t}dt$$
so
$$y=\dfrac{\ln^2{(1-x)}}{1-x}+\sum_{n=1}^{\infty}\left(1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}\right)x^n$$
then you can use:Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
A: Write down the function
$$ g(z) = \sum_{n\geq1} \frac{z^n}{n}H_n^2, $$
so that $S=g(-1)$ and $g$ can be reduced to
$$ zg'(z) = \sum_{n\geq1} z^n H_n^2 = h(z). $$
Now, using $H_n = H_{n-1} + \frac1n$ ($n\geq2$), we can get a closed form for $h(z)$:
$$h(z) = z + \sum_{n\geq2}\frac{z^n}{n^2} + \sum_{n\geq 2}z^n H_{n-1}^2 + \sum_{n\geq 2} 2\frac{z^n}{n}H_{n-1}. $$
Now, the first and third sums Mathematica can evaluate itself in closed form (the third one evaluates to the function $p(z)$ below, the first one is $\text{Li}_2(z)-z$), and the middle sum is $z h(z)$.
Substituting this into the expression for $g(z)$, we get
$$g(z) = \int \frac{\text{Li}_2(z) + p(z)}{z(1-z)}\,dz, $$
$$p(z) = -\frac{\pi^2}{3} + 2\log^2(1-z)-2\log(1-z)\log(z)+2\text{Li}_2((1-z)^{-1}) - 2\text{Li}_2(z). $$
Mathematica can also evaluate this integral, giving (up to a constant of integration)
\begin{align}
g(z) &= \frac{1}{3} \left(-2 \log(1-z^3+3 \log(1-z)^2 \log(-z)+\log(-1+z)^2 (\log(-1+z)+3 \log(-z) \right. \\ 
& \hspace{5mm} \left. -3 \log(z))+\pi ^2 (\log(-z)-2 \log(z))+\log(1-z) \left(\pi^2 - 3 \log(-1+z)^2 \right. \right.\\ 
& \hspace{5mm} \left.\left. +6 (\log(-1+z)-\log(-z)) \log(z)\right)-6 (\log(-1+z)-\log(z)) \left(\text{Li}_{2}\left(\frac{1}{1-z}\right)-\text{Li}_{2}(z)\right) \right.\\ 
& \hspace{10mm} \left. -3 \log(1-z) \text{Li}_{2}(z)+3 \text{Li}_{3}(z)\right).
\end{align}
The constant of integration is fixed by requiring $g(0)=0$.
Some care needs to be taken, because the function
is multi-valued, when evaluating $g(-1)$. The answer is
$$ \frac{1}{12}(\pi^2\log2-4(\log 2)^3-9\zeta(3)). $$
A: we have
$$ \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$
replace $x$ with $-x$, divide both sides by $x$ then integrate w.r.t $x$ from $0$ to $1$ , we get:
\begin{align*}
S_1&=\sum_{n=1}^{\infty}(-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ dx=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\frac{(-1)^n}n=\underbrace{\int_0^1\frac{\ln^2(1+x)}{x(1+x)}\ dx}_{x=\frac{1-y}{y}}\\
&=\int_{1/2}^1 \frac{\ln^2x}{1-x}\ dx=\sum_{n=1}^{\infty}\int_{1/2}^1x^{n-1}\ln^2x\ dx=\sum_{n=1}^{\infty}\left(\frac{2}{n^3}-\frac{2}{2^n n^3}-\frac{2\ln2}{2^n n^2}-\frac{\ln^22}{2^n n}\right)\\
&=2\zeta(3)-2\operatorname{Li_3}\left(\frac12\right)-2\ln2\operatorname{Li_2}\left(\frac12\right)-\ln^32
\end{align*}
Now using the identity:
$$\displaystyle \int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac1n\left(H_n^2+H_n^{(2)}\right)$$
multiply both sides by $(-1)^n$ then sum both sides w.r.t $n$ from $1$ to $\infty$, we get
\begin{align*}
S_2&=\sum_{n=1}^{\infty}\left(H_n^2+H_n^{(2)}\right)\frac{(-1)^n}{n}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^{\infty}(-x)^n\ dx=\underbrace{-\int_0^1\frac{\ln^2(1-x)}{1+x}\ dx}_{x=1-y}\\
&=-\int_0^1\frac{\ln^2(x)}{2-x}=-\sum_{n=1}^{\infty}\frac1{2^n}\int_0^1 x^{n-1}\ln^2x\ dx=-2\sum_{n=1}^{\infty}\frac1{2^n n^3}=-2\operatorname{Li_3}\left(\frac12\right)
\end{align*}
we are now ready to calcualate our sum:
\begin{align*}
\frac{S_1+S_2}{2}=\sum\frac{(-1)^n H_n^2}{n}&=\zeta(3)-2\operatorname{Li_3}\left(\frac12\right)-\ln2\operatorname{Li_2}\left(\frac12\right)-\frac12\ln^32\\
&=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32
\end{align*}
and as a bonus :
\begin{align*}
\frac{S_2-S_1}{2}=\sum\frac{(-1)^n H_n^{(2)}}{n}&=\ln2\operatorname{Li_2}\left(\frac12\right)-\zeta(3)+\frac12\ln^32\\
&=\frac12\ln2\zeta(2)-\zeta(3)
\end{align*}
where the results of $\operatorname{Li_3}\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$ and $ \operatorname{Li_2}\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$ were used in the calculations. 
A: Here is a solution using simple tools
We have
$$\sum_{n=1}^\infty x^nH_n=-\frac{\ln(1-x)}{1-x}$$
Replace $x$ with $-x$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{x(1+x)}\ dx$$
$$=\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{-5/8\zeta(3)}-\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ dx}_{\frac{1}{1+x}=y}$$
$$=-\frac58\zeta(3)-\int_{1/2}^1\frac{\ln\left(\frac{y}{2y-1}\right)\ln y}{y}\ dy=-\frac58\zeta(3)-I$$
$$I=\int_{1/2}^1\frac{\ln^2y}{y}\ dy-\int_{1/2}^1\frac{\ln(2y-1)\ln y}{y}\ dy=\frac13\ln^32-\Re\int_{1/2}^1\frac{\ln(1-2y)\ln y}{y}\ dy$$
$$=\frac13\ln^32+\Re\sum_{n=1}^\infty \frac{2^n}{n}\int_{1/2}^1 y^{n-1}\ln y\ dy=\frac13\ln^32+\Re\sum_{n=1}^\infty\frac{2^n}{n}\left(\frac{\ln2}{n2^n}+\frac{1}{n^22^n}-\frac{1}{n^2}\right)$$
$$=\frac13\ln^32+\ln2\zeta(2)+\zeta(3)-\Re\text{Li}_3(2)=\frac18\zeta(3)-\frac12\ln2\zeta(2)+\frac13\ln^32$$
where we used $\Re\text{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)$
Plug the result of $I$ we get $$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32$$

A different way to find $\int\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx$
First, add and subtract $\ln2$ and note that $\int\frac{\ln\left(\frac{1-x}{2}\right)}{1+x}\ dx=-\text{Li}_2\left(\frac{1+x}{2}\right)$
$$\int\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\int\frac{\ln\left(\frac{1-x}{2}\right)\ln(1+x)}{1+x} \ dx+\ln2\int\frac{\ln(1+x)}{1+x}\ dx$$
$$\overset{IBP}{=}-\ln(1+x)\text{Li}_2\left(\frac{1+x}{2}\right)+\int\frac{\text{Li}_2\left(\frac{1+x}{2}\right)}{1+x}\ dx+\frac12\ln2\ln^2(1+x)$$
$$=-\ln(1+x)\text{Li}_2\left(\frac{1+x}{2}\right)+\text{Li}_3\left(\frac{1+x}{2}\right)+\frac12\ln2\ln^2(1+x)$$
Therefore
$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$
A: Or we can use the generating function
$$\sum_{n=1}^\infty\frac{H_{n}^2}{n}x^{n}=\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x)-\frac13\ln^3(1-x)$$
By setting  $x=-1$ we get
$$\sum_{n=1}^\infty\frac{H_n^2}{n}(-1)^n=-\frac34\zeta(3)+\frac12\ln2\zeta(2)-\frac13\ln^32$$
note that $\operatorname{Li}_3(-1)=-\frac34\zeta(3)$ and $\operatorname{Li}_2(-1)=-\frac12\zeta(2)$
