Is the double series $\sum_{n,k\geq 1}\frac{1}{n^2+k^2}$ divergent?

Question

I thought that since the function $f(x,y)=(x^2+y^2)^{-1}$ is not integrable in $\mathbb{R}^2$ (I mean away from the origin) the same integral but with counting measure may behave the same. I tried this approach (similar to polar coordinates in some sense) $$ \sum_{n,k\geq 1}\frac{1}{n^2+k^2}=\sum_{t\geq1}\frac{r(t)}{t} $$ where $r(t)$ is the function that counts how many ways an integer $t$ can be written as sum of two squares. Does it exists any esteem or asymptotic behaviour for this function? Or maybe a different way to solve the problem?

Thanks in advance!

Answer 1

You're definitely on the exact right track: It's not summable for the exact same reason that $(x^2 + y^2)^{-1}$ is not integrable (and that basically gives a proof: the counting measure just looks like a discrete version of the integral, and by breaking up the plane into a union of squares of side-length $1$ this argument can be made very precise).

As far as your question on the asymptotics of $r$, see this question here.

Answer 2

Your approach is good, but there is a simpler solution.

$n^2+k^2\leq (n+k)^2$, so $\sum\limits_{n,k\geq 1} \frac{1}{n^2+k^2} \geq \sum\limits_{n,k\geq 1} \frac{1}{(n+k)^2}$.

Now a number $N$ can be written in $N-1$ ways as a sum $N=n+k$. So the lower estimate is $\sum\limits_{N\geq 1} \frac{N-1}{N^2}$. As $\sum\limits_{N\geq 1} \frac{1}{N^2}$ converges, the important part is $\sum\limits_{N\geq 1} \frac{N}{N^2} = \sum\limits_{N\geq 1} \frac{1}{N}=\infty$.