How to bound $\sum_{ \substack{\mathbf{x} \in \mathbb{Z}^n \\ \mathrm{dist}(\mathbf{x}, B) > \delta }} \frac{1}{\mathrm{dist}(\mathbf{x}, B)^m }$?

Question

I am interested in bounding the following sum $$\sum_{ \substack{\mathbf{x} \in \mathbb{Z}^n \\ \operatorname{dist}(\mathbf{x}, B) > \delta }} \frac{1}{\operatorname{dist}(\mathbf{x}, B)^m }, $$ where $B$ is a compact subset of $\mathbb{R}^n$ and $\operatorname{dist}(x, B) = \min_{\mathbf{y} \in B} \| \mathbf{x} - \mathbf{y} \|_{\infty}$ and $\delta > 0$. I am wondering for what values of $m$ will this sum converge, and how I can show it. Any comments would be appreciated. Thank you.

Answer 1

Here are some thoughts.

First, translate $B$, and suppose without loss of generality that $B$ is centered at origin (I'm being very sloppy here, but stay with me). Since $B$ is compact, it is closed and bounded, and thus, there is an $R>0$ such that $\max_{x\in B}\|x\|_\infty\leqslant R$.

Now, note that, if $\|y\|_\infty>R+\delta$ then we automatically get ${\rm dist}(y,B)>\delta$. Indeed, let $x\in B$. If $\|y-x\|_\infty\leqslant \delta$, then by triangle inequality, $\|y\|_\infty \leqslant \|y-x\|_\infty+\|x\|_\infty\leqslant R+\delta$. Thus, for every $x\in B$, $\|x-y\|_\infty\geqslant \delta$, if $\|y\|\geqslant R+\delta$.

I'll now vaguely approximate the sum you want to understand with: $$ \sum_{y\in\mathbb{Z}^n : \|y\|_\infty>R+\delta} \frac{1}{\|y-R\|_\infty^m}. $$ Now, $\|y-R\|_\infty = t$ implies, $R-t\leqslant y_i\leqslant R+t$ for every $t$. There are roughly $(2t)^n$ such vectors. Hence, the sum you're studying behaves like $$ \sum_{t\in\mathbb{Z}^+} \frac{(2t)^n}{t^m} = 2^n\sum_{t\in \mathbb{Z}^+}\frac{1}{t^{m-n}}. $$ Since the dimension $n$ is fixed, I'd expect that the threshold is near $m>n+1$.