2
$\begingroup$

Why is it that $$\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n}$$ diverges while $$\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n} + \frac{1}{n}$$ converges?

These are both the series representations of $$ \pi \cot(\pi z) $$ derive in Ahlfors page 189. The terms in the second series can be written so that one can compare the series with $\sum 1 / n^2$ rather than $\sum 1 / n$ and hence it converges.

Here's where I'm unsure. If I'm not mistaken, the second series is simply obtained from the first by adding and subtracting $1/n$ for all $n \in \mathbb{N}$ from the negative and positive sides of the series. Therefore, in my mind, the series should be equal so how can it be that the second series converges while the first does not?

My intuitive guess is that it has to do with the rate at which we are adding terms to the series but I don't know how to formalize that.

$\endgroup$
6
  • $\begingroup$ What will happen if $z=0$? Try to develop this idea. $\endgroup$
    – Jihad
    May 6, 2016 at 6:18
  • 2
    $\begingroup$ The first series is a representation of $\pi \cot(\pi z)$ only if it's interpreted as a Cauchy principal value (i.e., $\lim_{N \to \infty} \sum_{n=-N}^{N} \frac{1}{z-n}$). $\endgroup$
    – Random Variable
    May 6, 2016 at 11:05
  • $\begingroup$ Ok, I edited the summations to account for the singularity at the origin, but my question remains. $\endgroup$
    – mathjacks
    May 6, 2016 at 17:13
  • 1
    $\begingroup$ nothing complicated here. $\displaystyle\frac{1}{z-n} + \frac{1}{n} = \frac{n + z-n}{(z-n)n} = \frac{z}{nz-n^2}$ and thanks to the $n^2$ : $\quad\displaystyle z\sum_{n > |z|} \frac{1}{nz-n^2}$ converges. the same kind of calculation shows that $\displaystyle\sum_{n=1}^\infty \frac{1}{z-n} + \frac{1}{z+n}$ converges (whenever $z \not \in \mathbb{Z}$) $\endgroup$
    – reuns
    May 6, 2016 at 17:15
  • $\begingroup$ @user1952009. Thanks. I understand how to show the second series converges but my question is, if that is true, then why does the first series not converge? Are they not equivalent? I think you can get the second series by adding and subtracting $\frac{1}{n}$ from the first, which doesn't change its value. So why doesn't the first series converge? $\endgroup$
    – mathjacks
    May 6, 2016 at 18:01

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .