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 $n-1$.

\begin{align}
   \sum_{n=5}^\infty \frac{8}{n^2 -1}
   &=\sum_{n=5}^\infty \frac{8}{(n -1)(n+1)}\\
   &=\sum_{n=5}^\infty \left(\frac{4}{n-1}-\frac{4}{n+1} \right)\\
   &=\left(\frac 44 - \color{red}{\frac 46} \right)
   + \left(\frac 45 - \color{blue}{\frac 47} \right)
   + \left(\color{red}{\frac 46} - \color{green}{\frac 48} \right)
   + \left(\color{blue}{\frac 47} - \frac 49 \right)
   + \left(\color{green}{\frac 48} - \frac{4}{10} \right) \dots \\
   & =\frac 44 +\frac{4}{5} \\
   &=\frac{9}{5}
\end{align} 
