TITLE: Convergence of double sum $\sum_{m, n}\frac{1}{m^p + n^k}$
QUESTION [7 upvotes]: For which ${k, p \in \mathbb{R}}$ does $\sum_{m,n \in \mathbb{N}} \frac{1}{m^p + n^k}$ converge?
It is necessary but not sufficient that $p, k \gt 1$
I am looking for a simple solution, as the solutions I have seen so far have been rather perverted.

REPLY [3 votes]: Here is another approach:
Upper bound the sum by grouping by dyadic scales:
$\sum_{m,n} \frac{1}{m^k+n^p}\leq \sum_{l} \#\{m,n:\ 2^{l-1}\leq m^k+n^p \leq 2^l \} 2^{-(l-1)}$
We can upper bound the size of this set by the size of the set $\{m,n:\  m^k \leq 2^l, n^p \leq 2^l\}$, Thus the size is bounded above by $2^{l/k+l/p}$ and an upper bound on the sum is
$\sum_l 2^{1/k+1/p-1}$ which is finite if $1/k+1/p < 1$.
For the other direction, lower bound the sum by
$\sum_{m,n} \frac{1}{m^k+n^p}\geq \sum_{l} \#\{m,n:\ 2^{l-1}\leq m^k+n^p \leq 2^l \} 2^{-l}$
A lower bound on this count is to bound the set $\{m,n:\ 2^{l-2}\leq m^k \leq 2^{l-1}, 2^{l-2} \leq n^p \leq 2^{l-1}\}$ which has size of order $(2^{-1/k}-2^{-2/k})(2^{-1/p}-2^{-2/p})2^{l/k+l/p}$, and thus if $1/k+1/p > 1$ the sum diverges.