1
$\begingroup$

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 $a > 2$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $a > 2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.