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I know that the series converges. My questions is to what. I tried seeing if it was a telescoping series: $\sum_{n=2}^\infty \frac{2}{n^3-n} = 2\sum_{n=2}^\infty (\frac{1}{n^2-1}-\frac{1}{n})$ but it doesn't seem to cancel any terms. Thoughts?

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    $\begingroup$ Note that $\frac{1}{n^2-1}=\frac{1}{2} \left(\frac{1}{n-1}-\frac{1}{n+1}\right)$ $\endgroup$ – Mark Apr 11 '16 at 20:07
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That partial fraction is incorrect.

$\frac1{n^3-n} =\frac1{n(n^2-1)} =\frac1{n(n-1)(n+1)} =\frac{a}{n}+\frac{b}{n-1}+\frac{c}{n+1} $.

Find $a, b, $ and $c$ and then see if things cancel out in the sum.

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It is telescoping series: the term is $\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}$, so all the terms cancel except the initial $-\frac{2}{2}+\frac{1}{1}+\frac{1}{2}=\frac{1}{2}$.

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  • $\begingroup$ @Cipra All my signs the wrong way around. Now fixed. Many thanks. $\endgroup$ – almagest Apr 11 '16 at 20:24

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