I am working through the problems on this page. I am stuck on the fourth problem. In the hints they claim that $\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}=\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$. But why does this equality hold?
We have: $$\frac{1}{k-1}+\frac{1}{k+1} =\frac{k+1}{(k-1)(k+1)}+\frac{k-1}{(k+1)(k-1)}=\frac{(k+1)+(k-1)}{(k-1)(k+1)}=\frac{2k}{(k-1)(k+1)}=\frac{2k}{k^2-1}\neq \frac{k}{k^2-1} $$
And therefore shouldn't we have $\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}\neq\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$?
This is likely just a typo, in any case it doesn't change the problem much at all because this is a dead end: you can't evaluate the sum this way because it diverges.