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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

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    $\begingroup$ u may write this as the derivative of a Hurwitz eta function which can be reudced to usal zeta function and their derivative. $\endgroup$ – tired Dec 22 '15 at 1:47
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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.

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Notice that we have the Dirichlet eta function:

$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$

And its derivative:

$$\eta'(s)=-\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n)}{n^s}$$

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