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I tried to calculate the sum of the following series without success, any clue would be helpful! $$\sum_{n=2}^\infty \ln\left(1+\frac{(-1)^n}{n}\right) $$

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  • $\begingroup$ Can you tell us what you have tried so far so that we can help you continue from there? $\endgroup$ – Jesse P Francis Oct 5 '15 at 11:47
  • $\begingroup$ A sum of log is the log of product, and the product is rather easy to compute, just try a few terms. $\endgroup$ – Jean-Claude Arbaut Oct 5 '15 at 11:51
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Hint: $$\prod_{n=2}^{2N}\left(1+\frac{(-1)^n}{n}\right)=\prod_{m=1}^{N}\left(1+\frac{1}{2m}\right)\prod_{m=1}^{N-1}\left(1-\frac{1}{2m+1}\right)=\frac{2N+1}{2N}=1+\frac{1}{2N}. $$

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    $\begingroup$ And funnily, for odd products, you get exactly $1$, which is easily seen by multiplying your formula by $1-1/(2N+1)=2N/(2N+1)$. $\endgroup$ – Jean-Claude Arbaut Oct 5 '15 at 11:59

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