Convergence of $\sum_{n=1}^\infty \sum_{m=1}^\infty (n^\alpha+m^\beta)^{-s}$ Let $\alpha,\beta>0$.
My question: What is the infimum $s_0$ of all $s>0$ such that
$$\sum_{n=1}^\infty \sum_{m=1}^\infty (n^\alpha+m^\beta)^{-s}$$
converges?
Upper bound: I've found out that $s_0 \le \max(2/\alpha,2/\beta)$:
We have
$$n^\alpha+m^\beta \ge 2 n^{\alpha/2}m^{\beta/2}.$$
Hence,
$$\sum_{n=1}^\infty \sum_{m=1}^\infty (n^\alpha+m^\beta)^{-s} \le 2^{-s} \sum_{n=1}^\infty n^{-s\alpha/2} \sum_{m=1}^\infty n^{-s\beta/2} = 2^{-s}  \zeta(s\alpha/2)\zeta(s\beta/2).$$
So for the latter series to converge we must have $s>2/\alpha$ and $s>2/\beta$.
Lower bound: We have
$$\sum_{n=1}^\infty \sum_{m=1}^\infty (n^\alpha+m^\beta)^{-s} > \sum_{n=1}^\infty (n^\alpha+1)^{-s} \ge \sum_{n=1}^\infty (n^\alpha+n^\alpha)^{-s} = 2^{-s} \zeta(s\alpha).$$
So we get $\max(1/\alpha,1/\beta) \le s_0$.
 A: One way of approaching this would be to introduce
$$ r_{\alpha, \beta}(x) = r(x) = \# \{(n,m) \in \Bbb N : n^\alpha + m^\beta = x\}$$
and denoting $X = \{x \in \Bbb R : r(x) \geq 1\}.$
For sufficiently large $s$, this directly gives
$$ \sum_{n,m \in \Bbb N} (n^\alpha + m^\beta)^{-s} = \sum_{x \in X} r(x) x^{-s} $$
Note that $X \cap [1,N] \ll \#\{(n,m) : \max(n^\alpha, m^\beta) \leq N\} \ll N^{\frac 1\alpha + \frac 1\beta},$ and we also have the trivial bound $r(x) \ll x^{\min(1/\alpha, 1/\beta)}.$ Using these bounds, we obtain
$$\sum_{x \in X \ \ N/2 < x \leq N}r(x) x^{-s} \ll N^{\gamma - s}  \quad \text{for } \gamma = \frac 1\alpha + \frac 1\beta + \min(1/\alpha, 1/\beta),$$ which directly gives absolute convergence of the series for $\Re s > \gamma.$
However, we can even bound the prior sum assuming $r(x) = 1$ for every $x \in X$ at the cost of $\varepsilon$ in the exponent! We do this by, after being given a pair $N, \epsilon > 0$, choosing $\alpha - \varepsilon < \alpha' < \alpha$ and $ \beta - \varepsilon < \beta' < \beta$ such that $r_{\alpha', \beta'}(x) = 1$ for every $x \in X_{\alpha', \beta'} \cap [1,N]$.
The details are a bit tedious, but I can evaluate if there is need.
For every $\varepsilon > 0,$ this gives the improved bound
$$\sum_{x \in X \ \ N/2 < x \leq N}r(x) x^{-s} \ll N^{\gamma - s + \varepsilon} \quad \text{for } \gamma = \frac 1\alpha + \frac 1\beta,$$
which gives absolute convergence for any $s \in \Bbb C$ with $\Re s > \frac 1\alpha + \frac 1\beta.$
This is optimal at least for $\alpha = \beta = 1,2.$ I would expect this to be optimal in any case, though.
A: The series
$$\sum_{n=1}^\infty \sum_{m=1}^\infty (n^\alpha+m^\beta)^{-s}$$
is related to the integral
$$\int_1^\infty \int_1^\infty (x^\alpha+y^\beta)^{-s} dx dy.$$
Using substitution one sees for $a>0$
$$\int_0^\infty (x^\alpha+a)^{-s} dx = a^{\frac{1}{\alpha}-s} \int_0^\infty (x^\alpha+1)^{-s} dx.$$
The latter integral exists iff $s>1/\alpha$. Let's say
$$C_s := \int_0^\infty (x^\alpha+1)^{-s} dx.$$
So in the case $s>1/\alpha$ we have
$$\int_1^\infty \int_1^\infty (x^\alpha+y^\beta)^{-s} dx dy
\le \int_1^\infty \int_0^\infty (x^\alpha+y^\beta)^{-s} dx dy\\
= C_s\int_1^\infty (y^\beta)^{\frac{1}{\alpha}-s} dy.$$
This integral converges iff
$$\beta(\frac{1}{\alpha}-s)<-1$$
which is equivalent to
$$s > \frac{1}{\alpha}+\frac{1}{\beta}.$$
