How do I evaluate this sum :$ \sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}\log n}{n^s}$?

Question

How do I evaluate this sum :$$ \sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}\log n}{n^s}$$

Note : In wolfram alpha it is convergent for $Re(s)>1$ .!!

Thank you for any help

Answer 1

To expand on @tired's comment, your sum is

$$ \zeta'(s)\left(2^{1-s}-1\right)-\zeta(s)2^{1-s}\ln2 $$

Though if you simply want to compute it numerically, you may also consider the conceptually simpler act of computing successive partial sums until convergence up to a desired error tolerance.