# The limit of an alternating series [duplicate]

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

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

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 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 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

## marked as duplicate by user17762, Charles, EricMay 23 '12 at 20:46

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: $$\sum_{k=1}^\infty (-1)^k \frac{\log(k)}{k} = \gamma \log(2) - \frac{\log^2(2)}{2}$$