<s>[INST] Solving Telescoping Series $$\ \sum_{n=5}^\infty \frac{8}{n^2-1} $$
I tried the following:
$$\ \sum_{n=5}^\infty \frac{8}{n^2-1} = \sum_{n=5}^\infty \frac{8}{n-1} - \frac{8}{n} =$$  $$\left(2-\frac{8}{5}\right) + \left(\frac{8}{5} - \frac{8}{6}\right) + \left(\frac{8}{6} - \frac{8}{7}\right) + \cdots + \left(-\frac{8}{n}\right)$$
Terms cancelled each other out, therefore we are left with:
$$ \ (2 - \frac{8}{n}) $$
I could think the series converges to 2 since: 
$$\lim_{n \to \infty} \left(2-\frac 8 n \right) = 2$$ 
However, the correct answer is $$ \frac{9}{5} $$ what am I doing wrong?
 [/INST] Your partial fraction is wrong:
$$\frac{8}{n^2-1}=\frac{-4}{n+1}+\frac{4}{n-1}$$
Using similar trick, you will see that most terms cancel out and you will left with $1+\frac45.$
Edit:
to obtain the partial fractions,
Since $n^2-1=(n-1)(n+1)$,
$$\frac{8}{(n-1)(n+1)}= \frac{A}{n+1}+\frac{B}{n-1}.$$
We can for instance equate the two and solve for $A$ and $B$ by comparing coefficients.
I use a trick call heaviside cover method.
To determinte $A$, $n+1=0$, $n=-1$. I will cover up the term $n+1$ in the denominator of the left hand side and evalute $\frac{8}{n-1}$ with $n=-1$, hence $A=-4$. 
We can do similar stuff for $B$ as well. 
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