In terms of finding the sum of the series $\sum_{n=5}^{\infty} \frac{6}{n^2 - 3n}$ , this series looks to me like it is telescoping. So, I tried to factor it in a way to find a telescoping pattern, but I was unable to.

  • 1
    $\begingroup$ Have you tried partial fractions, then telescoping? $\endgroup$
    – Kenny Wong
    Jun 1, 2018 at 21:10
  • $\begingroup$ This is indeed telesoping through partial fraction decomposition (PFD first of course). Why don't you show us what you did before a ton of people are going to answer this for you? $\endgroup$
    – imranfat
    Jun 1, 2018 at 21:10

1 Answer 1


$\dfrac{6}{n(n-3)} = \dfrac{A}{n}+\dfrac{B}{n-3}$

$A(n-3) +Bn = 6$

$A = -2, B = 2$

This gives you:

$$\sum_{n=5}^\infty \left( \dfrac{2}{n-3}-\dfrac{2}{n} \right)$$

  • $\begingroup$ $=2\left(\sum_{n=2}^{\infty}\frac{1}{n}-\sum_{n=5}^{\infty}\frac{1}{n}\right)=2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)$ $\endgroup$
    – Cesareo
    Jun 1, 2018 at 22:52

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