# Why is $\sum_{k=2}^{\infty} \frac{k}{k^{2}-1}=\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$?

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 $$\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}\neq\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$$?

• You're right: $\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}=\frac12\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$ Jan 2, 2021 at 11:50
• Doesn't it diverge? Jan 3, 2021 at 12:50
• yes it does, but my question only concerned the equivalence of the two sides. But like the others said, it is probably a typo - the equivalence is wrong Jan 3, 2021 at 14:33