<s>[INST] 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 $S$?
 [/INST] 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. 
</s>