How to show the series $\sum_{\xi\in\mathbb Z^n}\langle \xi\rangle^{2s}$ converges if and only if $s<-n/2$?

Question

I need some help to show the following result: The "series", $$\displaystyle \sum_{\xi\in\mathbb Z^n}\langle \xi\rangle^{2s},$$ converges if and only if $s<-n/2$. The main problem is that this series is indexed in $\mathbb Z^n$, I don't know how to prove convergence in this case.. Here $\langle \xi\rangle=(1+|\xi|^2)^{1/2}$ and $|\xi|$ is the Euclidian norm of $\xi$. Thanks

Answer 1

If $s\geqslant0$, the series diverges. If $s\lt0$, using the identity $$ \langle\xi\rangle^{2s}=(-2s)\int\mathbf 1_{t\geqslant\langle\xi\rangle}t^{2s-1}\mathrm dt, $$ one sees that the series is $$ (-2s)\int_0^\infty t^{2s-1}\cdot\#A(t)\cdot\mathrm dt, $$ where $A(t)=\{\xi\in\mathbb Z^d\mid\langle\xi\rangle\leqslant t\}$ for every positive $t$. When $t\to+\infty$, $\#A(t)$ behaves like $t^n$ in the sense that for every $t$ large enough, $$ a\cdot t^n\leqslant\#A(t)\leqslant a'\cdot t^n, $$ hence the integral above converges if and only if $$ \int_1^\infty t^{n+2s-1}\cdot\mathrm dt $$ converges, that is, if and only if $n+2s-1\lt-1$.