Convergence of $\sum_{x\in \mathbb{Z}^{d}}\frac{1}{1+|x|^{2}}$

Question

I'm trying to prove whether the following series converges $$\sum_{x\in \mathbb{Z}^{d}}\frac{1}{1+|x|^{2}}$$ where $|x| := x_{1}^{2}+\cdots+x_{d}^{2}$. My approach was: $$\sum_{x\in \mathbb{Z}^{d}}\frac{1}{1+|x|^{2}} \le \int_{\mathbb{R}^{d}}\frac{1}{1+|x|^{2}}dx = \frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\int_{-\infty}^{\infty}\frac{1}{1+r^{2}} <+\infty$$ where I used change of variables to $n$-dimensional polar coordinates. Is this correct? Any alternative approaches?

Answer 1

Fix $m\ge1.$ Then

$$\sum_{n=1}^{\infty}\frac{1}{m^2+n^2} \ge \int_1^\infty \frac{1}{m^2+x^2}\, dx = \int_{1/m}^\infty \frac{1}{m}\frac{1}{1+y^2}\, dy \ge \frac{1}{m}\cdot \frac{\pi}{4}.$$

Thus

$$\tag 1 \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2+n^2} \ge \sum_{m=1}^{\infty} \frac{1}{m}\cdot \frac{\pi}{4} =\infty.$$

For every $d\ge 2$ the sum $\sum_{x\in \mathbb{Z}^{d}}\dfrac{1}{1+|x|^{2}}$ contains copies of the sum on the left of $(1).$ Thus your series diverges for all $d\ge 2.$

Answer 2

There are $\sim r^d$ lattice points with $|x|<r$ and $\sim r^{d-1}$ lattice points with $r-1<|x|<r$. Hence when ordered and grouped by length, the sum should behave like $\sum_r\frac{r^{d-1}}{1+r^2}$, i.e., it won't converge when $d\ge 2$.