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My question is about part (b). The question is as follows:

Let $t>0$ be given and fixed, and define $F(z)$ by

$\displaystyle F(z) = \prod_{n=1}^\infty (1-e^{-2\pi n t}e^{2\pi i z})$

(b) Prove that $F$ vanishes exactly when $z = -int +m$ for $n\geq 1$ and $m \in \mathbb{Z}$. Then, explain why if $z_n$ is an enumeration of these zeros then we have that $\displaystyle \sum \frac{1}{|z_n|^2}= \infty \;\;\;and\;\;\;\sum \frac{1}{|z_n|^{2+\epsilon}}<\infty$ for any $\epsilon >0$.

Proving the first part wasn't too bad. I found that $F(z)=0$ exactly when $z = -int +m$. I'm having a hard time figuring out the second part. How do I show that

$\displaystyle \sum_{n=1}^\infty \sum_{m\in \mathbb{Z}}\frac{1}{m^2 +n^2 t^2} =\infty \;\;\;and\;\;\; \sum_{n=1}^\infty \sum_{m\in \mathbb{Z}}\frac{1}{(m^2+n^2t^2)^{1+\epsilon/2}}<\infty$

It may be helpful to note that this exercise is in the chapter about Hadamard's theorem.

Bumblebee
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slowspider
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  • See here. – Jean Marie Nov 16 '21 at 04:11
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    Note that $m^2+t^2n^2 \ge c(m^2+n^2)$ where $c=\max(t^2,1)$ so the series $\sum_{n=1}^\infty \sum_{m\in \mathbb{Z}}\frac{1}{(m^2+n^2t^2)^a}$ is majorized by $K\sum_{n=1}^\infty \sum_{m\in \mathbb{Z}}\frac{1}{(m^2+n^2)^a}$ which is easily shown convergent for $a > 1$ by using the integral test and polar coordinates – Conrad Nov 16 '21 at 05:57
  • Nice, I think that works! – slowspider Nov 16 '21 at 23:15

1 Answers1

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For the divergent series, $$\sum_{n \ge 1} \sum_{m \in \mathbb{Z}} \frac{1}{m^2 + n^2 t^2} \ge \sum_{n \ge 1} \sum_{m=1}^n \frac{1}{m^2 + n^2 t^2} \ge \sum_{n \ge 1} \frac{n}{n^2 + n^2 t^2} = \frac{1}{1+t^2} \sum_{n \ge 1} \frac{1}{n} = \infty.$$

angryavian
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