<s>[INST] Calculate $\sum_{k=2}^\infty \frac{1}{k^2 - 1}$ I'm wondering if someone could check on this working: 
$$\sum_{k=2}^\infty \frac{1}{k^2 - 1} = \sum_{k=2}^\infty \frac{1}{(k - 1)(k + 1)} = \frac{1}{2}\sum_{k=2}^\infty \frac{1}{k-1} - \frac{1}{k+1}$$
This is a nice telescoping thing that has a $k$-th partial sum that looks like this:
$$\frac{1}{1} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + ... + \frac{1}{k-1} - \frac{1}{k+1} = 1 + \frac{1}{2} - \frac{1}{k+1} - \frac{1}{k+1}$$
Skipping some formalities we find:
$$\sum_{k=2}^\infty \frac{1}{k^2 - 1} = \frac{1}{2} . \lim_{k \rightarrow \infty}\frac{3}{2}-\frac{2}{k+1} = \frac{1}{2} . \frac{3}{2} = \frac{3}{4}$$
 [/INST] You have
$$
...= 1 + \frac{1}{2} - \frac{1}{k+1} - \frac{1}{k+1}
$$
and there should be
$$
= 1 + \frac{1}{2} - \frac{1}{\color{cyan}{k}} - \frac{1}{k+1}
$$
but it is easy to correct and even with this mistake does not change the result.
</s>