Absolute convergence of $\sum_{m \geq 1 , n\geq 1} \frac{1}{m^\alpha+n^\beta}$ where $\alpha,\beta > 2$

Question

Show that this double sum is absolutely convergent. [$m,n \in \mathbb{C}$]

$$ \sum\limits_{\substack{m \geq 1 \\ n\geq 1}} \frac{1}{m^\alpha+n^\beta}, \quad\alpha,\beta > 2$$

Since we have that $$\left| \frac{1}{m^\alpha+n^\beta}\right|= \frac{1}{m^\alpha+n^\beta} < \frac{1}{m^2+n^2}<\frac{1}{m^2}$$, and because $\sum\limits_{\substack{m \geq 1}} \frac{1}{m^2}$ is convergent, can we then conclude by the Weierstrass M-test that $\sum\limits_{\substack{m \geq 1 \\ n\geq 1}} \left|\frac{1}{m^\alpha+n^\beta}\right|$ converges?

That only shows that $\sum_{m \geq 1} \frac{1}{m^\alpha + n^\beta}$ converges for each fixed $n$, but says nothing about the double sum. – A correct proof is for example here: https://math.stackexchange.com/questions/1800159/for-which-values-of-real-alpha-beta-does-sum-n-m-ge-1-frac1n-alp, or here: https://math.stackexchange.com/questions/520854/convergence-of-double-sum-sum-m-n-frac1mp-nk. — Martin R, Jul 12 '19 at 20:25
Does this answer your question? For which values of real $\alpha, \beta$ does $\sum_{n,m \ge 1} \frac{1}{n^{\alpha}+ m^{\beta}}$ converge? — Anne Bauval, Oct 30 '23 at 14:07

score 1 · Answer 1 · answered Jul 17 '19 at 02:48

Assuming you meant $m,n\in \mathbb N:$ Because $1/(m^\alpha + x^\beta)$ is decreasing in $x,$ we have

$$\sum_{n=1}^{\infty}\frac{1}{m^\alpha + n^\beta} \le \int_0^\infty \frac{1}{m^\alpha + x^\beta}\,dx.$$

Let $x=m^{\alpha/\beta}y$ to see the last integral equals

$$m^{\alpha/\beta-\alpha}\int_0^\infty \frac{1}{1 + y^\beta}\,dy.$$

Because $\beta > 2,$ the last integral converges.

Now $\alpha/\beta-\alpha<-1.$ Thus

$$\sum_{m=1}^{\infty}m^{\alpha/\beta-\alpha} <\infty.$$

This shows the double sum converges.

Absolute convergence of $\sum_{m \geq 1 , n\geq 1} \frac{1}{m^\alpha+n^\beta}$ where $\alpha,\beta > 2$

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