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How to find the value this sum converges to?$$\sum_{n\geq2}^{\infty}(-1)^{n+1}\frac{n}{n^2-1} $$ I've tried writing it like this $$\sum_{n\geq2}^{\infty}(-1)^{n+1}·n·\Bigg(\frac{1/2}{n-1}-\frac{1/2}{n+1}\Bigg) $$ and writing a few terms, but they won't cancel and I ended up with no ideas, any hint? I haven't learnt integration nor differentiation FYI.

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  • $\begingroup$ Ok I miss wrote one thing, thank guys. $\endgroup$ – iggykimi Jan 24 at 16:58
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Note that$$(-1)^{n+1}\frac n{n^2-1}=\frac12\times\frac{(-1)^{n+1}}{n-1}+\frac12\times\frac{(-1)^{n+1}}{n+1}.$$But$$\sum_{n=2}^\infty\frac{(-1)^{n+1}}{n-1}=\sum_{n=1}^\infty\frac{(-1)^n}n=-\log(2)$$and$$\sum_{n=2}^\infty\frac{(-1)^{n+1}}{n+1}=\sum_{n=3}^\infty\frac{(-1)^n}n=\left(\sum_{n=1}^\infty\frac{(-1)^n}n\right)+1-\frac12=-\log(2)+\frac12.$$

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Try writing $$\begin{align}n\left({1\over n-1}-{1\over n+1}\right)&= n\left({1\over n-1}-{1\over n} + {1\over n} -{1\over n+1}\right)\\ &=n\left({1\over n(n-1)}+{1\over n(n+1)}\right)\\&={1\over n-1}+{1\over n+1}\end{align}$$

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