0
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ 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, 2015 at 1:47

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Notice that we have the Dirichlet eta function:

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

And its derivative:

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .