How many points $\xi\in\mathbb Z^n$ are there satisfying $p\leq|\xi|< p+1$? Let $p\in\mathbb N$. How many points $\xi\in\mathbb Z^n$ are there satisfying $p\leq|\xi|< p+1$? Here $|\xi|$ indicates the usual Euclidian norm.
I am trying to decide convergence in How to show the series $\displaystyle\sum_{\xi\in\mathbb Z^n}\frac{1}{(1+|\xi|^2)^{s/2}}$ converges if and only if $s>n$?
 A: Denote by $X_{n,p}$ the number of the integer points $\xi\in\mathbb{Z}^n$ satisfying $p\le|\xi|<p+1$.
Here is a very rough upper bound for $X_{n,p}$.
For every $\xi=(\xi_1,\ldots,\xi_n)\in\mathbb{Z}^n$ denote by $E_\xi$ the unit cube translated to $\xi$:
$$
E_\xi := \xi + [0,1)^n = [\xi_1,\xi_1+1)\times\cdots\times[\xi_n,\xi_n+1).
$$
If $\xi,\eta\in\mathbb{Z}^n$ and $\xi\ne\eta$, then $E_\xi$ and $E_\eta$ are disjoint. If $A$ is a finite subset of $\mathbb{Z}^n$, then the number of elements of $A$ is equal to the $n$-volume of the union $\bigcup_{\xi\in A}E_\xi$.
The diameter of $E_\xi$ is equal to $\sqrt{n}$.
If $p>\sqrt{n}$ and $p\le|\xi|<p+1$, then $E_\xi$ is contained in the spherical shell $B(p+1+\sqrt{n})\setminus B(p-\sqrt{n})$.
Therefore $X_{n,p}$ is less than or equal to the $n$-volume of this shell:
$$
X_{n,p}
\le\omega_n((p+1+\sqrt{n})^n-(p-\sqrt{n})^n).
$$
In fact, the inequality is strict, but it is not important.
Apply the formula $a^n-b^n=(a-b)\sum_{j=0}^{n-1}a^j b^{n-1-j}$:
$$
X_{n,p}\le\omega_n (1+2\sqrt{n}) \sum_{j=0}^{n-1}(p+1+\sqrt{n})^j(p-\sqrt{n})^{n-1-j}.
$$
Estimate $p-\sqrt{n}$ by $p+1+\sqrt{n}$:
$$
X_{n,p}\le n\omega_n (1+2\sqrt{n}) (p+1+\sqrt{n})^{n-1}.
$$
This is enough to conclude that the original serie converges for every $s>n$.
