1
$\begingroup$

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

$\endgroup$
3
  • 5
    $\begingroup$ You're right: $\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}=\frac12\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$ $\endgroup$
    – A. Goodier
    Jan 2, 2021 at 11:50
  • $\begingroup$ Doesn't it diverge? $\endgroup$ Jan 3, 2021 at 12:50
  • $\begingroup$ 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 $\endgroup$
    – timtam
    Jan 3, 2021 at 14:33

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .