Does this reciprocal sum of an integer lattice converge and how do you show that?

Question

Let $(a,b) \neq 1$. Then $\sum\limits_{m, n \in \Bbb{Z}} \dfrac{1}{(am + bn)^2 - 1}$

I know that sum of $1/n^2$ converges, but this seems to have more terms.

Answer 1

As noted by Lord Shark the Unknown, the series diverges.

For $\gcd(a,b)\neq 1$ and a given $(m_0,n_0)\in\Bbb Z^2$, there are infinitely many integers $m,n$ such that $am+bn=am_0+bn_0=K\textrm{ (say)}$

Proof: Since $\gcd(a,b)=d\neq 1$, there exist integers $m,n\in\Bbb Z$ such that $am+bn=d$. In fact, there exist infinitely many such pairs. If $(m,n)$ is a pair of Bezout coefficients, then $(m+\frac{kb}d,n-\frac{ka}d)$ is another pair of Bezout coefficients where $k\in\Bbb Z$ is arbitrary (can you show this rigorously?)

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Hence, the sum contains a countable sum $\sum\limits_{n\in\Bbb Z}\frac{1}{K^2-1}$ which is divergent since $\frac{n}{K^2-1}\to\infty$ as $n\to\infty$. Hence, the original bigger sum diverges by comparison.

Answer 2

Intuitively, just like the sum of $\frac 1n$ diverges because each increment of length contributes almost as much as the last one, your sum is over all the lattice points in two dimensions and a ring contributes almost as much as the last one. You should expect that the sum diverges.

Slightly less informally, you can imagine that in a certain radius $r$ from the origin, an annulus of thickness $\Delta r$ has an area $2\pi r \Delta r$. You would expect $\frac {2\pi r \Delta r}{ab}$ points of your lattice in this area. Write that as a sum over $r$ to infinity, convert to an integral, and it diverges.