Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
The sum of $(-1)^n \frac{\ln n}{n}$

Compute $$\sum_{k=2}^{\infty}\frac{(-1)^{k}{\ln{k}}}{k}$$

share|cite|improve this question

marked as duplicate by Marvis, Charles, Eric Naslund May 23 '12 at 20:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
What is "n" in the log's argument? – DonAntonio May 23 '12 at 17:24
    
@EricNaslund I was just wondering why you didn't close this since it is an exact duplicate as you have pointed out in your comment. – user17762 May 23 '12 at 20:34
    
@Marvis: Done. In the past, there was some discussion about whether or not moderators should use their binding votes to close exact duplicates. (Notably several meta threads) I was just making sure at least one or two other member of the community agreed with closing. – Eric Naslund May 23 '12 at 20:49
    
@EricNaslund Ok. Yes it makes sense that the moderators wait for at-least another person to vote for closing a question. – user17762 May 23 '12 at 20:50
up vote 7 down vote accepted

Consider, for $s>1$, the following auxiliary convergent series: $$ \sum_{k=1}^\infty (-1)^k \frac{\log(k)}{k^s} = -\frac{\mathrm{d}}{\mathrm{d} s} \sum_{k=1}^\infty (-1)^k \frac{1}{k^s}= -\frac{\mathrm{d}}{\mathrm{d} s}\left( (2^{1-s} - 1)\zeta(s) \right) $$ The value of the series in question is obtained as a limit: $$ \sum_{k=1}^\infty (-1)^k \frac{\log(k)}{k} = \lim_{s \searrow 1} \left( 2^{1-s} \log(2) \zeta(s) + \zeta^\prime(s) (1-2^{1-s}) \right) $$ Since $\zeta(s) = \frac{1}{s-1} + \gamma + \mathcal{O}(s-1)$, and $\zeta^\prime(s) = -\frac{1}{(s-1)^2} + \mathcal{O}(1)$ we arrive at: $$ \sum_{k=1}^\infty (-1)^k \frac{\log(k)}{k} = \gamma \log(2) - \frac{\log^2(2)}{2} $$

share|cite|improve this answer
    
thanks for your answer. – user 1618033 May 23 '12 at 19:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.