For which values of real $\alpha, \beta$ does $\sum_{n,m \ge 1} \frac{1}{n^{\alpha}+ m^{\beta}}$ converge? I was wondering how does the series $$\sum_{n,m \ge 1} \frac{1}{n^{\alpha}+ m^{\beta}}$$
behave for real $\alpha, \beta > 0$.
My approach: firstly I considered the case $\alpha = \beta > 2$. Then
$$\lim_{m \to +\infty} \sum_{n \ge 1} \frac{m^{\alpha-1}}{n^{\alpha}+ m^{\alpha }} = \lim_{m \to +\infty} \frac{1}{m} \sum_{n \ge 1} \frac{1}{(n/m)^{\alpha}+1} = \int_0^{+ \infty} \frac{1}{x^{\alpha}+1} \mathrm{d} x = C < \infty$$
so that
$$ \sum_{m=1}^{+ \infty} \sum_{n=1}^{+ \infty} \frac{1}{n^{\alpha}+ m^{\alpha }} = \sum_{m=1}^{+ \infty} O \left( \frac{1}{m^{\alpha -1}} \right)$$
is convergent.
Using a similar method, I proved that if $\alpha = \beta \le 2$ then the series is divergent.
However, this method doesn't work for $\alpha \neq \beta$, and the only thing I managed to conclude is that if both $\alpha, \beta > 2$, then we have convergence, if both $\alpha , \beta \le 2$ then we have divergence.
How can we study the case $\alpha \le 2 < \beta$?
 A: You can view the sum as the integral over $[1,\infty) \times [1,\infty)$ of the function whose value on $[m,m+1) \times [n,n+1)$ is ${\displaystyle {1 \over m^{\beta} + n^{\alpha}}}$. This integrand is within a fixed factor of ${\displaystyle {1 \over x^{\beta} + y^{\alpha}}}$, so what you are asking is equivalent to determining the convergence of
$$\int_1^{\infty}\int_1^{\infty} {dx\,dy \over x^{\beta} + y^{\alpha}}$$
Let $x = X^{2 \over \beta}$ and $y = Y^{2 \over \alpha}$. Then this becomes
$${4 \over \alpha \beta}\int_1^{\infty}\int_1^{\infty} { x^{{2 \over \beta}- 1}y^{{2 \over \alpha} - 1}\over x^2 + y^2}\,dx\,dy$$
Converting to polar this turns into an integral of the form
$${4 \over \alpha \beta}\int_D r^{{2 \over \alpha } + {2 \over \beta} - 3}
\cos(\theta)^{{2 \over \beta} - 1}\sin(\theta)^{{2 \over \alpha} - 1}\,dr\,d\theta$$
Because $\cos(\theta)^{{2 \over \beta} - 1}\sin(\theta)^{{2 \over \alpha} - 1}$ is integrable, in view of the shape of the domain, the integral above will converge if and only if the power of $r$ appearing is less than $-1$. 
So the condition is that ${\displaystyle {2 \over \alpha } + {2 \over \beta} - 3 < - 1}$, which is equivalent to $${1 \over \alpha} + {1 \over \beta} < 1$$
A: Let $R(N)$ be the number of solutions $(n,m)\in\mathbb{N}^+\times\mathbb{N}^+$ of $n^{\alpha}+m^{\beta}\leq N$. We have:
$$ R(N)\approx \sum_{k=1}^{N^{1/\alpha}}(N-k^{\alpha})^{1/\beta}=N^{1/\beta}\sum_{k=1}^{N^{1/\alpha}}\left[1-\left(\frac{k}{N^{1/\alpha}}\right)^{\alpha}\right]^{1/\beta}$$
that by Riemann sums behaves like:
$$ R(N) \approx N^{1/\beta+1/\alpha}\int_{0}^{1}(1-x^{\alpha})^{1/\beta}\,dx = N^{1/\alpha+1/\beta}\cdot \frac{\Gamma\left(1+\frac{1}{\alpha}\right)\Gamma\left(1+\frac{1}{\beta}\right)}{\Gamma\left(1+\frac{1}{\alpha}+\frac{1}{\beta}\right)}.$$
It follows that $R(N+k)-R(N)$ behaves like $N^{1/\alpha+1/\beta-1}$, but by condensation the convergence/divergence of the original series just depends on the convergence/divergence of the series
$$ \sum_{N\geq N_0}\frac{R(N+k)-R(N)}{N} \sim \sum_{N\geq N_0}\frac{R(N+k)-R(N)}{N+k}. $$
However, the series
$$ \sum_{N\geq N_0}\frac{N^{\frac{1}{\alpha}+\frac{1}{\beta}-1}}{N} $$
is convergent iff 

$$\frac{1}{\alpha}+\frac{1}{\beta}< 1. $$

We may see the last one is a sufficient condition also by reindexing the original double series, then applying Hilbert's inequality ($\ell^p$ version).
