# Determining Convergence of Double Series by Comparison

I’m trying to determine for which values of $$a >1$$ there is convergence of the double series $$\sum_{(n,m)\in \mathbb{N}^2}\frac{1}{m^a+n^a}$$. One possible approach is to use the integral test checking convergence of $$\int_1^\infty \int_1^\infty \frac{dxdy}{x^a+ y^a}$$, but I want to try this with a comparison. I think could show that it diverges if $$a \leq 2$$ with the iterated sum:

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^a + n^a} > \sum_{m=1}^\infty \sum_{n=1}^m \frac{1}{m^a + n^a}> \sum_{m=1}^\infty \frac{m}{2m^a}= \sum_{m=1}^\infty \frac{1}{2m^{a-1}}$$

The series on the right side diverges when $$a \leq 2$$

My questions are: (1) Does proving divergence this way with an iterated sum prove divergence of the double series? and (2) How could I use a comparison test to prove convergence or divergence for $$a > 2$$?

Since the terms are nonnegative, the double series converges if and only if the iterated series converges. Your approach proving divergence when $$a \leqslant 2$$ is valid.
More generally, a double series with nonnegative terms can be summed in any way, e.g., along diagonals. This can be used to prove convergence here for all $$a >2$$.
Since $$x \mapsto x^a$$ is convex, we have $$\frac{1}{2} (m^a + n^a) \geqslant \left(\frac{m+n}{2} \right)^a$$ which implies $$\frac{1}{m^a + n^a} \leqslant \frac{2^{a-1}}{(m+n)^a}$$
$$\sum_{m,n =1}^\infty\frac{1}{m^a+n^a} = \sum_{q= 2}^\infty \sum_{m+n=q}\frac{1}{m^a+n^a} \leqslant 2^{a-1}\sum_{q=2}^\infty\sum_{m+n =q}\frac{1}{q^a}\\ = 2^{a-1}\sum_{q=2}^\infty\frac{q-1}{q^a}$$
The series on the RHS (and, hence, the double series on the LHS) converges for all $$a > 2$$.