Convergence of double series involving minimum 
Determine the convergence of the series $$\sum_{n,m\in\mathbb{Z}:|n-m|>10,|m-10|>0}\min\{|n|^{-10},|n-m|^{-10}\}.$$

I tried solving this using "an integral test", saying
$$
\sum_{n,m\in\mathbb{Z}:|n-m|>10,|m-10|>0}\min\{|n|^{-10},|n-m|^{-10}\} \le \int_{|x-y|>10,|y-10|>0} \min \{|x|^{-10},|x-y|^{-10}\}dxdy
.$$
However the integral to the right diverges (it's bounded from below by $\int_{-\infty}^{-10}\int_{20}^\infty \min\{...\} dxdy=\iint_{[20,\infty)\times[-\infty,-10)}|x|^{-10}dxdy$ which is divergent). It does not mean however anything as the sum may still be convergent.
Does anybody has an idea how to show this series converges/diverges?
 A: 1. Let me first convince you that the sum converges by investigating a much easier version:
$$ \sum_{(n,m)\neq 0} |n|^{-10}\wedge|m|^{-10} $$
Decomposing the sum over the subregions divided by the lines $m = \pm n$, and exploiting the symmetry, this sum is bounded from above by
$$ 4 \sum_{n=1}^{\infty} \sum_{m=-n}^{n} \frac{1}{n^{10}} = 4 \sum_{n=1}^{\infty} \frac{2n+1}{n^{10}}, $$
which converges.

2. Now we move on to the sum in the question. Since the sum $\sum_{|n-10|>10} |n|^{-10} \wedge |n-10|^{-10} $ converges, it suffices to study the convergence of
$$ S := \sum_{|n-m|>10} |n|^{-10} \wedge |n-m|^{-10}. $$
By noting that
$$ |n-m| \geq |n| \quad \Longleftrightarrow \quad (m \geq 0 \text{ and } n \leq 2m) \text{ or } (m \leq 0 \text{ and } n \geq 2m), $$
$\hspace{80pt}$
we get
\begin{align*}
S
&\leq 2 \sum_{\substack{|n-m|>10 \\ m \geq 0}} |n|^{-10} \wedge |n-m|^{-10} \\
&\leq 2 \sum_{\substack{|n-m|>10 \\ m \geq 0 \\ n \geq 2m}} |n|^{-10} + 2 \sum_{\substack{|n-m|>10 \\ m \geq 0 \\ n \leq 2m}} |n-m|^{-10} \\
&\leq 2 \sum_{m \geq 0} \sum_{\substack{n \geq m/2 \\ n \neq 0}} |n|^{-10} + 2\sum_{l=11}^{\infty} \frac{\#\{ (n,m) : m=n+l, m \geq 0, n \leq 2m\}}{l^{10}}.
\end{align*}
Now by using the fact that
$$ \sum_{\substack{n \geq m/2 \\ n \neq 0}} |n|^{-10} = \mathcal{O}(m^{-9}) $$
and
$$ \#\{ (n,m) : m=n+l, m \geq 0, n \leq 2m\} = \mathcal{O}(l), $$
it follows that the above upper bound converges.
