# Find the sum of the series: $\sum_{n=5}^{\infty} \frac{6}{n^2 - 3n}$

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.

• Have you tried partial fractions, then telescoping? – Kenny Wong Jun 1 '18 at 21:10
• 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? – imranfat Jun 1 '18 at 21:10

$\dfrac{6}{n(n-3)} = \dfrac{A}{n}+\dfrac{B}{n-3}$
$A(n-3) +Bn = 6$
$A = -2, B = 2$
$$\sum_{n=5}^\infty \left( \dfrac{2}{n-3}-\dfrac{2}{n} \right)$$
• $=2\left(\sum_{n=2}^{\infty}\frac{1}{n}-\sum_{n=5}^{\infty}\frac{1}{n}\right)=2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)$ – Cesareo Jun 1 '18 at 22:52