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 $n-1$.
 A: Looking a little closer at the question, he is asking about partial fraction decomposition, as opposed to the value of the sum itself. For this particular example, it's fairly straight forward.
When given a fraction which contains a polynomial denominator, you can factor this fraction and break it into a sum of other fractions with denominators of a lower polynomial order.
For example, we begin be identifying that:
$$\frac{8}{n^2 - 1} = \frac{8}{(n+1)(n-1)}$$
We are then interested in finding factors $A,\ B$ such that
$$\frac{8}{n^2 - 1} = \frac A{n+1} + \frac B{n-1}$$
All we need to do now is combine the right side, and solve for the necessary $A,\ B$, to equate to the right side.
$$\frac A{n+1} + \frac B{n-1} = \frac{A(n-1) + B(n+1)}{(n+1)(n-1)} = \frac{(A+B)n + B - A}{(n+1)(n-1)}$$
Since your original fraction has $8$ in the numerator, we make the following comparison:
$$(A+B)n + B - A = 8$$
To conclude that
$$A+B = 0;\ \ \ B - A = 8\implies -A = B = 4$$
Hopefully this gets you past where you are stuck, and the supplementary answers can take you the rest of the way!
EDIT:
The $(n-1)^{-1}$ and $(n-3)^{-1}$ terms arise as part of the series. When considering this series as it pushes on to infinity, the observation of $(n-1)^{-1}$ and $(n-3)^{-1}$ is no different from looking at $(n+1)^{-1}$ and $(n-1)^{-1}$. What is important is the difference between them ($2$ indices).
A: Factor $n^2 - 1$ as a difference of squares. Then you will end up with the fraction $\frac{8}{(n-1)(n+1)}$. You can now use partial fractions to solve the problem.
A: The correct way to analyze this is to write
$$\begin{align}
\sum_{n=5}^N\frac{2}{n^2-1}&=\sum_{n=5}^{N}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\\\\
&=\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right)+\left(\frac16-\frac18\right)+\cdots \\\\
&+\left(\frac1{N-3}-\frac1{N-1}\right)+\left(\frac1{N-2}-\frac1N\right)+\left(\frac1{N-1}-\frac1{N+1}\right)\\\\
&=\frac14+\frac15+\frac{1}{N}+\frac{1}{N+1}
\end{align}$$
Therefore,
$$\sum_{n=5}\frac{8}{n^2-1}=4\lim_{N\to \infty}\left(\frac14+\frac15+\frac{1}{N}+\frac{1}{N+1}\right)=\frac95$$
A: $$S = \sum_{n=5}^\infty \frac{8}{n^2 -1}=
\lim_{N\to\infty} \sum_{n=5}^N \frac{8}{n^2 -1}
= \lim_{N\to\infty} \sum_{n=5}^N \left\{ {4\over n-1} - {4\over n+1}\right\} = \lim_{N\to\infty} \sum_{n=5}^N {4\over n-1} - \sum_{n=5}^N {4\over n+1}$$
Now reindex to get 
$$S =  \lim_{N\to\infty} \sum_{n=4}^{N-1} {4\over n} - \sum_{n=6}^{N+1} {4\over n} = \lim_{N\to\infty} 1 + {4\over 5} - {4\over N}- {4\over N+1} = {9\over 5}$$
A: \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} 
