How to show the series $\displaystyle\sum_{\xi\in\mathbb Z^n}\frac{1}{(1+|\xi|^2)^{s/2}}$ converges if and only if $s>n$? Let $s\in\mathbb R$. How to show the series $$\sum_{\xi\in\mathbb Z^n}\frac{1}{(1+|\xi|^2)^{s/2}}$$ converges if and only if $s>n$ ($n$ is the dimension of $\mathbb Z^n$)? The convergence of this series is to be understood as the existence of the limit $$\lim_{k\to \infty}\sum_{|\xi|\leq k}\frac{1}{(1+|\xi|^2)^{s/2}}.$$ Above $|\xi|$ indicates the usual Euclidian norm. 
 A: Using standard methods, we get
$$
\begin{align}
\int_{\mathbb{R}^n}\frac{\mathrm{d}x}{\left(1+|x|^2\right)^{s/2}}
&=n\omega_n\int_0^\infty\frac{r^{n-1}\,\mathrm{d}r}{\left(1+r^2\right)^{s/2}}\\
&=\frac{n}{2}\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}\int_0^\infty\frac{r^{n/2-1}\,\mathrm{d}r}{\left(1+r\right)^{s/2}}\\
&=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)}\frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{s-n}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\\
&=\pi^{n/2}\frac{\Gamma\left(\frac{s-n}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\tag{1}
\end{align}
$$
The integral in $(1)$ is decreasing in $s$, and as $s$ decreases to $n$, the integral grows like
$$
\frac{c_n}{s-n}\tag{2}
$$
where $c_n=\dfrac{2\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)}$. Thus, the integral in $(1)$ converges precisely when $s\gt n$.
Each point in the lattice sits at the center of its own unit cube with sides parallel to the axes. For the cube centered on the lattice point $k\in\mathbb{Z}^n$, when $|k|^2\ge n$,
$$
\frac14\le\frac{1+(|k|-\sqrt{n}/2)^2}{1+|k|^2}\le\frac{1+|x|^2}{1+|k|^2}\le\frac{1+(|k|+\sqrt{n}/2)^2}{1+|k|^2}\le\frac94\tag{3}
$$
Thus,
$$
\sum_{k\in\mathbb{Z}^n}\frac1{\left(1+|k|^2\right)^s}\tag{4}
$$
converges precisely when the integral in $(1)$ converges; that is, when $s\gt n$.
A: Here is not an answer, just some hints.
First way. Try to compare the original sum with the integral
$$
\int\limits_{x\in\mathbb{R}^n}\frac{dx_1\cdots dx_n}{(1+\|x\|^2)^{s/2}}.
$$
Start with $\int_{t\in\mathbb{R}}\frac{dt}{(a^2+t^2)^{s/2}}$, make a suitable change of variables and reduce it to the Beta function.
Second way. It is something like a "polar change of variables". Try to estimate the inner sum and reduce the original problem to a one-dimensional problem:
$$
\sum_{p=0}^{\infty}\sum_{10p\le|\xi|<10p+10}\frac{1}{(1+\|\xi\|^2)^{s/2}}.
$$
