Closed form of $\sum_{n=1}^{\infty}(-1)^n \frac{\ln^2(n)}{n}$ I am working a maths problem and came across this sum. I am unable to evaluates it.
$$\sum_{n=1}^{\infty}(-1)^n \frac{\ln^2(n)}{n}$$
I have checked on wolfram and wikipedia maths pages but can't seem to finds it.
Can anyone please point to its closed form? Thank you.
 A: According to a CAS
$$\sum_{n=1}^{\infty}(-1)^n \frac{\log^2(n)}{n}=2 \gamma _1 \log (2)-\frac{1}{3}\log ^3(2)+\gamma  \log ^2(2)\approx 0.0653726$$ where $\gamma _1$ is a Stieltjes constant.
If we use, as J.G. commented,
$$\eta(s)=\sum_{n=1}^\infty(-1)^{n-1}n^{-s}=\left(2^{1-s}-1\right) \zeta (s)$$
$$\frac{d^2}{ds^2}\eta(s)=2^{-s} \left(2 \log ^2(2)   \zeta (s)-4 \log (2) \zeta '(s)-\left(2^s-2\right) \zeta ''(s)\right)$$ and then the result.
Edit
Let us consider the general case
$$S_p =\sum _{n=1}^{\infty } (-1)^n\frac{ \log ^p(n)}{n}$$ The results are
$$\left(
\begin{array}{cc}
p & \frac{S_p}{\log(2)} \\
 1 & \gamma -\frac{\log (2)}{2} \\
 2 & \gamma  \log (2)-\frac{\log ^2(2)}{3}+2 \gamma _1 \\
 3 & \gamma  \log ^2(2)-\frac{\log ^3(2)}{4}+3\log (2) \gamma _1+3 \gamma _2 \\
 4 & \gamma  \log ^3(2)-\frac{\log ^4(2)}{5}+4 \log ^2(2) \gamma _1+6\log (2)
   \gamma _2+4 \gamma _3 \\
 5 & \gamma  \log ^4(2)-\frac{\log ^5(2)}{6}+5 \log ^3(2) \gamma _1+10 \log ^2(2)
   \gamma _2+10 \log (2) \gamma _3+5 \gamma _4 
\end{array}
\right)$$ where interesting patterns appear.
