Asymptotic behavior of $\sum_{k=0}^{N-1} \sum_{n \neq k} \frac{1}{\vert e_k^N-e_n^N \vert^2}$ as $N\to\infty$, where $e_k^N :=e^{2\pi i k/N}$

Question

Consider the unit circle $\{x \in \mathbb C: \vert x \vert_2=1 \}$ then we can now consider the roots of unity $$e_k^N :=e^{2\pi i k/N}\text{ for }k \in \{0,..,N-1\}$$ on that circle.

We can now define the sum over all possible inverse-square distances

$$\sum_{k=0}^{N-1} \sum_{n \neq k} \frac{1}{\vert e_k^N-e_n^N \vert^2}$$

My question is:

How does this sum behave as $N$ tends to infinity? Obviously, it will go to infinity, but what is the asymptotic behaviour?

Please let me know if you have any questions.

Answer 1

Let $e_k = e^{2 \pi i k/N}$. Since $$\sum_{0 \leq k < N \land k \neq l} |e_k - e_l|^{-2}$$ is the sum of inverse squares of the distances from the $l$th vertex of a regular $N$-gon to the other $N - 1$ vertices, it does not depend on $l$. The sum for $l = 0$ can be evaluated as $$\sum_{0 < k < N} |e_k - 1|^{-2} = \sum_{0 < k < N} \frac 1 4 \csc^2 \frac {\pi k} N = \sum_{0 < k < N} \, \operatorname* {Res}_{z = e_k} \frac N {(1 - z)^2 (1 - z^N)} = \\ -\operatorname* {Res}_{z = 1} \frac N {(1 - z)^2 (1 - z^N)} = \frac {N^2 - 1} {12}.$$

Answer 2

Since $|e^{ix}-e^{iy}|^2=4\sin^2\frac{x-y}{2}$, your sum is$$S_N:=\frac14\sum_{n\ne k}\csc^2\frac{\pi(k-n)}{N}\sim\frac{N^2}{4}\int_{\Omega_N}dxdy\csc^2\pi(x-y),$$with $\Omega_N\subseteq[0,\,1]^2$ reproducing the off-diagonal behaviour of the double sum. This nuance ensures the integral itself is finite for any finite $N$, but grows with increasing $N$, so the $S_N$ diverges faster than $N^2$. It won't diverge faster than $N^4$, however, as it's a sum of $O(N^2)$ terms, each with a $O(N^2)$ upper bound.

Numerical experiments, suggest $S_N\sim cN^3$ with $c\approx\frac{1}{12}$. I'm not sure why $c$ would take that value, but we can justify a $O(N)$ behaviour for the integral by noting $\Omega_N$ excises a width-$O(1/N)$ region near the origin; on its edges, the integrand is $O(N^2)$.