Convergence of Multidimensional Infinite Series

I am trying to analyze the convergence of multidimensional infinite sums such as those in the following form:

$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{1}{1+\alpha\exp (\beta n) \exp (\gamma m)}$$

where $\alpha,\beta,\gamma\in(0,\infty)$.

I'm quickly realizing though, that I have no experience with analyzing sums like this. I realize that because $\mathbb{N}^2$ is countable, I could in principle re-write this sum as a related one:

$\sum_{n=0}^{\infty} f(n)$

where $f(n)$ could even be guarenteed monotone decreasing. However, I've been having trouble getting anywhere with this.

Is there some well-established method of attacking convergence of multidimensional sums like this one? If so, I would very much appreciate being pointed in the right direction.

Solution 1:

The question of absolute convergence can be answered by studying how quickly the function decays as $(n, m) \in \mathbb{N}_{\ge 0}^2$ gets further away from the origin. For example, let $A_k = \{ (n, m) : n + m = k \}$. Then we can write the sum as

$$\sum_{k \ge 0} \sum_{(n, m) \in A_k} \frac{1}{1 + \alpha \exp (\beta n) \exp (\gamma m)}.$$

The inner sum is a sum over $k + 1$ terms, each of which is bounded above by $\frac{1}{1 + \alpha \exp (\text{min}(\beta, \gamma) k)}$, so the whole sum is bounded by

$$\sum_{k \ge 0} \frac{C_1 k}{\exp (C_2 k)}$$

for some constants $C_1, C_2$, and this obviously converges.

The basic idea is to reduce the question to a one-dimensional question in a natural way. It is closely related to, for example, determining whether a multidimensional integral absolutely converges by bounding its integral over larger and larger spheres.

The question of conditional convergence is much more subtle, since it strongly depends on what order you sum the terms in (for example, in general you can't switch the inner and outer sums), and unlike the one-dimensional case there isn't a privileged order.

Solution 2:

If you just want to know if this converges or not, the answer is that it converges since $\alpha>0$ implies $$\frac1{1 + \alpha \exp(\beta n) \exp(\gamma m)} < \frac1{\alpha \exp(\beta n) \exp(\gamma m)}$$

$$\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac1{1 + \alpha \exp(\beta n) \exp(\gamma m)} < \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac1{\alpha} \exp(-\beta n) \exp(-\gamma m)$$

It is a geometric series and since $\beta,\gamma > 0$ the series converges and we have $$\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac1{1 + \alpha \exp(\beta n) \exp(\gamma m)} < \frac1{\alpha} \frac1{1-\exp(-\beta)} \frac1{1 - \exp(-\gamma)}$$

Hence the series converges.