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

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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.

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

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

$A = -2, B = 2$

This gives you:

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