Convergence of double series over $\mathbb{Z}$ I am attempting to show that the following double series converges
\begin{equation}
\sum\limits_{a,b \in \mathbb{Z}} \frac{1}{\left(\sqrt{a^2 +b^2} - M\right)^3}
\end{equation}
where $M \in \mathbb{R}^{+}$ is a constant. I've never dealt with double series before, but this is what I have attempted so far:
First let's define $$u_{ab} := \frac{1}{\left(\sqrt{a^2 +b^2} - M\right)^3}  \hspace{10mm}\left(\text{$\textbf{Notice:}$ $u_{a,b} = u_{-a,b}=u_{a,-b}=u_{-a,-b}$}\right)$$
Then our double series becomes
\begin{align*}
\sum\limits_{a=-\infty}^{\infty}\left(\sum\limits_{b=-\infty}^{\infty} u_{ab}\right) &= \sum\limits_{a=-\infty}^{\infty}\left(u_{a0} + 2\sum\limits_{b=1}^{\infty} u_{ab}\right) &&\text{(by $\textbf{Notice}$)}\\[10pt] 
&= \sum\limits_{a=-\infty}^{\infty}u_{a0} + 2\sum\limits_{a=-\infty}^{\infty}\left(\sum\limits_{b=1}^{\infty} u_{ab}\right)\\[10pt]
&=u_{00} +2 \left(\sum\limits_{a=1}^{\infty}u_{a0}\right) + 2\left[\sum\limits_{b=1}^{\infty}u_{0b} + 2\sum\limits_{a=1}^{\infty}\sum\limits_{b=1}^{\infty}u_{ab}\right]\\[10pt]
&=u_{00} + 2\left(\sum\limits_{a=1}^{\infty} u_{a0} + \sum\limits_{b=1}^{\infty}u_{0b}\right) + 2\sum\limits_{a,b=1}^{\infty}u_{ab}\\[10pt]
&=u_{00} + 4\sum\limits_{a=1}^{\infty}u_{a0} + 2\sum\limits_{a,b=1}^{\infty}u_{ab} &&\left(\text{since $u_{ab}=u_{ba}$}\right)
\end{align*}
Firstly: is my analysis above correct?

Secondly: I know that $\sum\limits_{a=1}^{\infty}u_{a0}$ converges, however the latter series is a double series, I have no experience/knowledge on how to show convergence for this. Any help? Intuition points towards some sort of $2d$ integral test? But I'm very unsure about that.
 A: Your analysis so far looks good to me. I think it is worth stating explicitly in your question that $M$ is not equal to $\sqrt{a^2+b^2}$ for any integers $a,b$.
Now, with that out of the way: we have $a^2+b^2\ge 2ab$ for all $a,b$. So for $a\ge \frac{M^2}{(\sqrt 2-1)^2}$ or $b\ge \frac{M^2}{(\sqrt 2-1)^2}$, $|\sqrt{a^2+b^2}-M|\ge \sqrt{ab}$. So the sum over this region is bounded by
$$\sum_{a,b\ge 1} \left(\frac{1}{ab}\right)^\frac32=\sum_{a\ge 1} \left(\frac{1}{a}\right)^\frac32 \sum_{b\ge 1} \left(\frac{1}{b}\right)^\frac32$$
which converges.
A: Your series equals
$$ \sum_{n\geq 0}\frac{r_2(n)}{(\sqrt{n}-M)^3} \tag{1}$$
where
$$ r_2(n)=\left|\left\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\right\}\right| $$
equals
$$ 4\sum_{d\mid n}\chi_4(d),\qquad \chi_4(d)=\left\{\begin{array}{rcl}1&\text{if}&d\equiv 1\pmod{4}\\-1&\text{if}&d\equiv 3\pmod{4}\\ 0 & \text{if}&d\equiv 0\pmod{2}\end{array}\right.$$
due to the fact that $\mathbb{Z}[i]$ is an Euclidean domain, hence a UFD. By Gauss circle problem we have that
$$ S(N)=\sum_{n=0}^{N}r_2(N) = \pi N + O(N^{1/2}) \tag{2}$$
since the LHS counts the number of lattice points in the region $a^2+b^2\leq N$. By summation by parts
$$ \sum_{n=0}^{N}\frac{r_2(n)}{(\sqrt{n}-M)^3}=\underbrace{\frac{S(N)}{(\sqrt{N}-M)^3}}_{\to 0}+\sum_{n=0}^{N-1}S(N)\left[\frac{1}{(\sqrt{n}-M)^3}-\frac{1}{(\sqrt{n+1}-M)^3}\right] $$
and the RHS is clearly convergent since its main term behaves like $C\cdot n^{-3/2}$ for large values of $n$.
