How can one evaluate the following series:

Question

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

From $\sum_{n\geq 1}\frac{(-1)^n \ln n}{n}$ it has been answered that

$$ \sum_{k=1}^{\infty} (-1)^{k} \frac{\ln{k}}{k} = \space \gamma \cdot \ln{2} \space - \space \frac{{(\ln{2})^{2}}}{2} \space \space \space , $$

where $\gamma$ is the Euler-Mascheroni constant.

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In the answer given by Olivier, maybe try to differentiate (3) one more time? — , Jul 29 '18 at 18:44

score 1 · Accepted Answer · answered Jul 29 '18 at 18:51

1

Hint. From the standard relation $$\sum_{n\geq 1}\frac{(-1)^n }{n^s}=\left(\frac{1}{2^{s-1}}-1\right)\zeta(s)$$ one may differentiate twice and set $s \to 1^+$ using the Laurent series of the Riemann zeta function.

answered Jul 29 '18 at 18:51

Olivier Oloa

120,989

Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation. – Leonardo Bohac Jul 29 '18 at 19:27
@LeonardoBohac You are right, typo has been corrected. Thank you. – Olivier Oloa Jul 30 '18 at 08:25

How can one evaluate the following series:

1 Answers1