<s>[INST] Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1}$ and so on? Like where does the $n-3$ come from or the $n-1$.
 [/INST] $$S = \sum_{n=5}^\infty \frac{8}{n^2 -1}=
\lim_{N\to\infty} \sum_{n=5}^N \frac{8}{n^2 -1}
= \lim_{N\to\infty} \sum_{n=5}^N \left\{ {4\over n-1} - {4\over n+1}\right\} = \lim_{N\to\infty} \sum_{n=5}^N {4\over n-1} - \sum_{n=5}^N {4\over n+1}$$
Now reindex to get 
$$S =  \lim_{N\to\infty} \sum_{n=4}^{N-1} {4\over n} - \sum_{n=6}^{N+1} {4\over n} = \lim_{N\to\infty} 1 + {4\over 5} - {4\over N}- {4\over N+1} = {9\over 5}$$
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