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?

  • 4
    $\begingroup$ 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

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.


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 $-\frac{2}{2}+\frac{1}{1}+\frac{1}{2}=\frac{1}{2}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.