I have to compute a double infinite sum to within a given accuracy $\epsilon$. Let us say the sum is of the form $$\sum_{m\geq 1} \sum_{n\geq 1} \frac{a_{m,n}}{m^2 n^2 \max(m,n)},$$ where $|a_{m,n}|\leq 1$.

The naïve approach would be to sum for all $(m,n)$ within a box and then bound the tails. That is not really efficient: for a box of size $\sqrt{N}$, we would be computing $N$ terms, and the error bound would be on the order of $1/N$. It is much better to consider a more general region $U\subset \mathbb{Z}^+\times \mathbb{Z}^+$, compute $$\sum_{(m,n)\in U} \frac{a_{m,n}}{m^2 n^2 \max(m,n)},$$ and bound the tails (i.e., the contribution of the complement $U^c = (\mathbb{Z}^+\times \mathbb{Z}^+)\setminus U$). For instance, for the fairly natural choice $$U = \{(m,n): m^2 n^2 \max(m,n)\leq M\},$$ the sum over $(m,n)\in U^c$ is bounded by about $(10/3)/M^{3/5}$, whereas $|U|\leq 6 M^{2/5}$. We can thus obtain an error of size $C/N^{3/2}$ when computing $N$ terms, where $C = \frac{10}{3} \cdot 6^{3/2}$.

Here comes the surprising part: that $U$ is not optimal. If we choose, more generally, $$U_\alpha = \{(m,n): (m n)^{\alpha} \max(m,n)\leq M\},$$ we find that $\alpha \approx 7$ is about twice as good as $\alpha=2$ (that is, it gives a better $C$, and so we get about half the error for the same number of terms).

Is there a good reason? Is there an even better choice of $U$?