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I'm trying to show that the following series diverges ($t$ fixed): $$\sum_{n > 0} \sum_{m \in \mathbb{Z}} \frac{1}{m^2+n^2t^2}$$

I'm trying to find some way to compare it to a smaller series that diverges, but I'm not getting any results. Can someone point me in the right direction?

angryavian
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  • You could compare it to an integral to see that it diverges. – Daniel Fischer Nov 02 '13 at 16:28
  • @DanielFischer Do you mean something like $\int_0^\infty \int_0^\infty \frac{1}{x^2+y^2} \mathop{dx}\mathop{dy}$? – angryavian Nov 02 '13 at 16:34
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    Something along those lines. Since $m$ traverses all of $\mathbb{Z}$, having one of the integrals extend over $\mathbb{R}$ would be a closer match, but that's immaterial. However, you'd want one of the integrals have a lower bound of $1$ or so ($> 0$) to not produce a singularity in the origin, since for the series, the only problem is at infinity. Then see how you can interpret the sum as an approximation to the integral. – Daniel Fischer Nov 02 '13 at 16:39

2 Answers2

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I just found a way:

$$\sum_{n>0} \sum_{m \in \mathbb{Z}}\frac{1}{m^2+n^2} \ge \sum_{n > 0} \sum_{m=1}^n \frac{1}{m^2+n^2} \ge \sum_{n>0}\sum_{m=1}^n \frac{1}{2n^2}=\sum_{n>0}\frac{1}{2n}$$

angryavian
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You may use any of several methods, e.g., the residue theorem, to evaluate the inner sum exactly:

$$\sum_{m \in \mathbb{Z}} \frac{1}{m^2+n^2 t^2} = \frac{\pi}{n t} \frac{1+e^{-2 n t}}{1-e^{-2 n t}}$$

The summand over $n$, then, is asymptotically a constant times $1/n$ for large $n$; therefore, the series then may be shown to be bounded from below by

$$\frac{\pi}{t} \sum_{n>0} \frac{1}{n}$$

which diverges.

Ron Gordon
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  • Sorry, on what function did you use the residue theorem? – angryavian Nov 02 '13 at 17:35
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    By an application of the residue theorem, for certain functions $f$: $$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \operatorname*{Res}_{z=z_k} \pi \cot{(\pi z)} f(z)$$ where the $z_k$ are the non-integer poles of $f$. – Ron Gordon Nov 02 '13 at 18:41