I would like an asymptotic estimate of $$ \sum_{y \in \mathbb{Z}^d} \frac{1}{|y-a_1|^{d-1} \ldots |y-a_n|^{d-1}} $$ that does not involve any infinite summation. In order to lighten the notation, I use $|z|$ for the maximum between the absolute value of $z$ and $1$, and I'm interested in $d \ge 3$. In order to get an idea of the result I'm looking for, we can start with $n=2$: it's easy to see that $$ \sum_{y \in \mathbb{Z}^d} \frac{1}{|y|^{d-1} \ |y-a|^{d-1}} \lesssim \frac{1}{|a|^{d-2}}. $$ The reasoning goes by cutting the summation into the part where $|y| \le |a|/2$, the part where $|y-a| \le |a|/2$, and the rest. Now, for $n = 3$, one expects that the bound on $$ \sum_{y \in \mathbb{Z}^d} \frac{1}{|y|^{d-1} \ |y-a|^{d-1} \ |y-b|^{d-1}} $$ will be the sum of three terms, corresponding to the summation around $0$, $a$ and $b$ respectively, and that the rest can be absorbed by these. For instance, the contribution from the ball around the origin gives a contribution of $$ \frac{|a| \wedge |b|}{|a|^{d-1} \ |b|^{d-1}}, $$ and so on for the other two terms. I think I have a cumbersome proof that this is correct. The natural generalisation would be that $$ \sum_{y \in \mathbb{Z}^d} \frac{1}{|y-a_1|^{d-1} \ldots |y-a_n|^{d-1}} \lesssim \sum_{i = 1}^n \frac{\min_{k \neq i} |a_k - a_i|}{\prod_{k \neq i} |a_k - a_i|^{d-1}}. $$ I haven't managed to prove this for now, and my attempts have left me to wonder whether it is actually true.
EDIT: I would be happy with the statement that for every $\epsilon > 0$, $$ \sum_{y \in \mathbb{Z}^d} \frac{1}{|y-a_1|^{d-1} \ldots |y-a_n|^{d-1}} \lesssim \sum_{i = 1}^n \frac{\min_{k \neq i} |a_k - a_i|}{\prod_{k \neq i} |a_k - a_i|^{d-1-\epsilon}}. $$ This is what I proved for $n =3$.
