Is $\sum_{n,m \in \mathbb Z^2} e^{-\Vert n-m \Vert} \frac{1}{1+\Vert n \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}}$ summable? I would like to ask whether the expression
$$\sum_{n,m \in \mathbb Z^2} e^{-\Vert n-m \Vert} \frac{1}{1+\Vert n \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}}$$
is finite?
Intuitively, this should be the case as away from the diagonal $n=m$ the exponential is rapidly decaying and on the diagonal, this expression is summable, but I cannot make it rigorous. 
EDIT: The sum is over $n$ and $m$ both in $\mathbb Z^2.$
 A: Note first that everything is positive. 
Then
$$\sum_{n,m \in \mathbb Z^2} e^{-\Vert n-m \Vert} \frac{1}{1+\Vert n \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}} \stackrel{m-n=k}{===}\sum_{k \in \mathbb Z^2}\sum_{m \in \mathbb Z^2} e^{-\Vert k \Vert} \frac{1}{1+\Vert m+k \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}} \\\stackrel{m-n=k}{===}\sum_{k \in \mathbb Z^2}e^{-\Vert k \Vert} \sum_{m \in \mathbb Z^2} \frac{1}{1+\Vert m+k \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}} $$ 
Now, for each $k \in \mathbb Z^2$ you have by Cauchy- Schwartz (you have to prove this step by looking at the partial sums)
$$\left(\sum_{m \in \mathbb Z^2} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m+k \Vert^{1+\varepsilon}} \right)^2 \leq  \left( \sum_{m \in \mathbb Z^2} \frac{1}{(1+\Vert m\Vert^{1+\varepsilon})^2} \right)\left( \sum_{m \in \mathbb Z^2} \frac{1}{(1+\Vert m+k\Vert^{1+\varepsilon})^2} \right)\\=\left( \sum_{m \in \mathbb Z^2} \frac{1}{(1+\Vert m\Vert^{1+\varepsilon})^2} \right)^2 $$
Now, it is easy to argue that 
$$C:= \left( \sum_{m \in \mathbb Z^2} \frac{1}{(1+\Vert m\Vert^{1+\varepsilon})^2} \right)< \infty$$
Therefore,
$$\sum_{n,m \in \mathbb Z^2} e^{-\Vert n-m \Vert} \frac{1}{1+\Vert n \Vert^{1+\varepsilon}} \frac{1}{1+\Vert m \Vert^{1+\varepsilon}} \leq \sum_{k \in \mathbb Z^2}e^{-\Vert k \Vert} \cdot C <\infty  $$ 
