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.
2026-03-25 22:27:02.1774477622
Find the sum of the series: $\sum_{n=5}^{\infty} \frac{6}{n^2 - 3n}$
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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:
$$\sum_{n=5}^\infty \left( \dfrac{2}{n-3}-\dfrac{2}{n} \right)$$