How to compute the value of an infinite series

Question

How do I compute this series$$\sum_{n=3}^{\infty}\frac{4}{n^2 - 4}$$ I know I should consider the k-th partial sum, which is $$\sum_{n=3}^{k}\frac{1}{2n - 8}-\frac{1}{2n+8}$$ By observing which terms cancel and do not, the terms that don't cancel add up to $\frac{29}{40}$, but I do not know how to then determine the limit as k tends to infinity.

Answer 1

Observe that :$\dfrac{4}{n^2-4}=\dfrac{1}{n-2} - \dfrac{1}{n+2} = \left(\dfrac{1}{n-2} - \dfrac{1}{n-1}\right)+ \left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+\left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)$, and use telescope for the $4$ sums.

Answer 2

$$S=\sum_{n=3}^\infty\frac4{n^2-4}=\sum_{n=3}^\infty\frac4{(n+2)(n-2)}$$ you could then use partial fractions to split this up into two harmonic sums with limits, then add them together