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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 $a > 2$?

RRL
  • 90,707
WoodWorker
  • 1,587

1 Answers1

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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 $$\frac{1}{m^a + n^a} \leqslant \frac{2^{a-1}}{(m+n)^a}$$

Thus,

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

The series on the RHS (and, hence, the double series on the LHS) converges for all $a > 2$.

RRL
  • 90,707