Convergence of double sum on the lattice I am working with a commutator $T$ acting on the lattice $\ell^2(\mathbb{Z}^2;\mathbb{C})$, the function space made up by the basis elements
\begin{align}\left|\vec{x}\right>\,:\,\mathbb{Z}^2&\rightarrow \mathbb{C}\\
\vec{y}&\mapsto \delta_{\vec{x},\vec{y}}
\end{align}
I am studying a paper in which the authors claim a particular (double) sum is convergent. It is the following,
\begin{equation}
C_N\sum_{r_1\in \mathbb{Z}}\sum_{x_1\in \mathbb{Z}}\left(1+\frac{1}{2}\left(|x_1+r_1|+|x_1|\right)\right)^{-N}.\qquad (1)
\end{equation}
Here, $N$ can be any real number, to each $N$ a certain constant $C_N$ is fit. In the paper, the authors claim this sum to be convergent, but I have a hard time realising why that is the case. At first thought, for fixed $r_1'\in \mathbb{Z}$, e.g. for $N=2$, the inner sum will look very much like the sum $\sum_{x_1\in \mathbb{Z}}\frac{1}{|x_1|^2}$, which I now converge to a finite number. But if I sum over this finite number an infinite number of times, obviously it will not converge. Is it possible to pick a proper $N$ such that the double sum in (1) does indeed converge?

It converges for $N>2$. For this we can use the (stupid) estimate
$$ \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N} \leq \frac{1}{(1+\vert r_1 + x_1\vert)^{N/2}} \frac{1}{(1+\vert x_1 \vert)^{N/2}}.$$
Then we have
$$ \sum_{x_1, r_1 \in \mathbb{Z}} \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N}
\leq \sum_{x_1, r_1 \in \mathbb{Z}} \frac{1}{(1+\vert r_1 + x_1\vert)^{N/2}} \frac{1}{(1+\vert x_1 \vert)^{N/2}}
= \left(\sum_{k\in\mathbb{Z}} \frac{1}{(1+\vert k \vert)^{N/2}} \right)^2.$$
For the last equality we first fixed $x_1$ and summed over $r_1$ (shifting it by $x_1$). For $N>2$ we have $N/2>1$ and so the RHS does converge and hence so does the LHS.
We can also analyse it a bit more carefully to see that it does not converge for $1<N<2$ (clearly it does not converge for $N\leq 1$, not even the series over a single variable). For this we will use the integral test, which tells us
$$ \frac{N-1}{(2+\vert x_1 \vert)^{N-1}} = \int_1^\infty \frac{1}{(1+\vert x_1\vert +r)^N} dr \leq \sum_{r\in \mathbb{N}_{\geq 1}} \frac{1}{(1+\vert x_1 \vert + r)^N} . $$
Thus, we obtain
$$ \sum_{x_1\in \mathbb{Z}} \sum_{r_1\in \mathbb{Z}} \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N}
\geq \sum_{x_1\in \mathbb{Z}} \left(  2 \sum_{r_1\in \mathbb{N}} \frac{1}{(1+\vert x_1 + r_1 \vert + \vert x_1 \vert)^N} \right) 
\geq 2\sum_{x_1\in \mathbb{Z}} \frac{N-1}{(2+\vert x_1 \vert)^{N-1}}.$$
Using the integral again, one sees that the RHS blows up and so our original series does not converge.
