Convergence of double sum $\sum_{m, n}\frac{1}{m^p + n^k}$ For which ${k, p \in \mathbb{R}}$ does $\sum_{m,n \in \mathbb{N}} \frac{1}{m^p + n^k}$ converge?
It is necessary but not sufficient that $p, k \gt 1$
I am looking for a simple solution, as the solutions I have seen so far have been rather perverted.
 A: Breaking the sum into two parts we have
$$
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^p+n^k}
=\color{#C00000}{\sum_{m=1}^\infty\sum_{n=1}^{\left\lfloor m^{p/k}\right\rfloor}\frac1{m^p+n^k}}
+\color{#00A000}{\sum_{n=1}^\infty\sum_{m=1}^{\left\lceil n^{k/p}-1\right\rceil}\frac1{m^p+n^k}}\tag{1}
$$
For the red (left hand) sum, where $m^p\ge n^k$, we have
$$
\frac12\sum_{m=1}^\infty\sum_{n=1}^{\left\lfloor m^{p/k}\right\rfloor}\frac1{m^p}
\le\sum_{m=1}^\infty\sum_{n=1}^{\left\lfloor m^{p/k}\right\rfloor}\frac1{m^p+n^k}
\le\sum_{m=1}^\infty\sum_{n=1}^{\left\lfloor m^{p/k}\right\rfloor}\frac1{m^p}\tag{2}
$$
and
$$
\sum_{m=1}^\infty\sum_{n=1}^{\left\lfloor m^{p/k}\right\rfloor}\frac1{m^p}
=\sum_{m=1}^\infty\frac{\left\lfloor m^{p/k}\right\rfloor}{m^p}\tag{3}
$$
which converges iff $p\left(1-\frac1k\right)\gt1$, which is equivalent to $\frac1p+\frac1k\lt1$.
For the green (right hand) sum, where $n^k\gt m^p$, we have
$$
\frac12\sum_{n=1}^\infty\sum_{m=1}^{\left\lceil n^{k/p}-1\right\rceil}\frac1{n^k}
\le\sum_{n=1}^\infty\sum_{m=1}^{\left\lceil n^{k/p}-1\right\rceil}\frac1{m^p+n^k}
\le\sum_{n=1}^\infty\sum_{m=1}^{\left\lceil n^{k/p}-1\right\rceil}\frac1{n^k}\tag{4}
$$
and
$$
\sum_{n=1}^\infty\sum_{m=1}^{\left\lceil n^{k/p}-1\right\rceil}\frac1{n^k}
=\sum_{n=1}^\infty\frac{\left\lceil n^{k/p}-1\right\rceil}{n^k}\tag{5}
$$
which converges iff $k\left(1-\frac1p\right)\gt1$, which is equivalent to $\frac1p+\frac1k\lt1$.
Thus, $(1)$ converges iff $\frac1p+\frac1k\lt1$.
A: Let us assume that $p,k>1$. Denote $$a_{m,n}=\frac{1}{m^p+n^k},$$
and for every $m\ge 1$, denote
$$b_m=m^{-p(1-\frac{1}{k})}\quad\text{and}\quad S_m=\sum_{n=1}^\infty a_{m,n}<\infty.$$

Claim: There exist $0<C_1<C_2$, only dependent on $k$ and $p$, such that for every  $m\ge 1$,
  $$C_1\, b_m<S_m<C_2\, b_m.\tag{1}$$ 
  As a corollary, 
  $$\sum_{m,n=1}^\infty a_{m,n}=\sum_{m=1}^\infty S_m<\infty \iff \frac{1}{p}+\frac{1}{k}<1.\tag{2}$$

Proof: To prove $(1)$, we may suppose $m^\frac{p}{k}\ge 3$ for convenience. Write $S_m$ as $A_m+B_m$, where
$$A_m=\sum_{1\le n\le m^\frac{p}{k}}a_{m,n}\quad\text{and}\quad\quad B_m=\sum_{n>m^\frac{p}{k}}a_{m,n}.$$
Given $m^\frac{p}{k}\ge 3$, when $1\le n\le m^\frac{p}{k}$, 
$$\frac{m^{-p}}{2} \le a_{m,n}\le m^{-p}\Longrightarrow \frac{b_m}{6}= \frac{m^\frac{p}{k}}{3}\cdot\frac{m^{-p}}{2} \le A_m\le m^\frac{p}{k}\cdot m^{-p}=b_m\,; \tag{3}$$
when $n>m^\frac{p}{k}$,
$$\frac{n^{-k}}{2} \le a_{m,n}\le n^{-k}\Longrightarrow \frac{b_m}{2^k(k-1)}=\int_{2m^\frac{p}{k}}^\infty\frac{dx}{2x^k}\le B_m\le\int_{\frac{1}{2}m^\frac{p}{k}}^\infty\frac{dx}{x^k}= \frac{2^{k-1} b_m}{k-1}. \tag{4}$$
$(1)$ follows from $(3)$ and $(4)$ immediately; $(2)$ follows from $(1)$ and the fact that
$$\sum_{m=1}^\infty m^{-p(1-\frac{1}{k})}<\infty \iff p(1-\frac{1}{k})>1.$$ 
A: Here is another approach:
Upper bound the sum by grouping by dyadic scales:
$\sum_{m,n} \frac{1}{m^k+n^p}\leq \sum_{l} \#\{m,n:\ 2^{l-1}\leq m^k+n^p \leq 2^l \} 2^{-(l-1)}$
We can upper bound the size of this set by the size of the set $\{m,n:\  m^k \leq 2^l, n^p \leq 2^l\}$, Thus the size is bounded above by $2^{l/k+l/p}$ and an upper bound on the sum is
$\sum_l 2^{1/k+1/p-1}$ which is finite if $1/k+1/p < 1$.
For the other direction, lower bound the sum by
$\sum_{m,n} \frac{1}{m^k+n^p}\geq \sum_{l} \#\{m,n:\ 2^{l-1}\leq m^k+n^p \leq 2^l \} 2^{-l}$
A lower bound on this count is to bound the set $\{m,n:\ 2^{l-2}\leq m^k \leq 2^{l-1}, 2^{l-2} \leq n^p \leq 2^{l-1}\}$ which has size of order $(2^{-1/k}-2^{-2/k})(2^{-1/p}-2^{-2/p})2^{l/k+l/p}$, and thus if $1/k+1/p > 1$ the sum diverges.
A: We split the sum under consideration into the three sums of positive terms:
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac 1 {m^p+n^k}=\sum_{m=2}^\infty \sum_{n=2}^\infty \frac 1 {m^p+n^k}+\sum_{m=1}^\infty \frac 1 {m^p+1}+\sum_{n=2}^\infty \frac 1 {n^k+1}.$$ If $p>1$ and $k>1$, then the two last series converge.
Next, $$ \sum_{m=2}^\infty \sum_{n=2}^\infty \frac 1 {m^p+n^k}\le \sum_{m=2}^\infty \sum_{n=2}^\infty \int_{m-1}^{m} \int_{n-1}^{n} \frac {dx\,dy} {x^p+y^k}=\int_1^\infty \int_1^\infty \frac {dx\, dy} {x^p+y^k}.$$
Now we change $$ \left\{ x={r}^{2/p} \left( \cos \left( \phi \right)  \right)
^{2/p},y={r}^{2/k} \left( \sin \left( \phi \right)
 \right) ^{2/k} \right\}.
 $$
 The Maple code $$  with(VectorCalculus):$$
$$simplify(Jacobian([r^{2/p}*cos(phi)^{2/p}, r^{2/k}*sin(phi)^{2/k}], [r, phi], 'determinant')[2], trig, power);$$ calculates the Jacobian $$4\,{r}^{-{\frac {pk-2\,k-2\,p}{pk}}} \left( \cos \left( \phi \right)
 \right) ^{-{\frac {p-2}{p}}} \left( \sin \left( \phi \right)
 \right) ^{-{\frac {-2+k}{k}}}{p}^{-1}{k}^{-1}.
 $$
 Thus, $$ \int_1^\infty \int_1^\infty \frac {dx\, dy} {x^p+y^k}\le$$ $$\int_1^\infty r^{-{\frac {pk-2\,k-2\,p}{pk}}-2} \,dr\int_{\phi_1(r)}^{\phi_2(r)}\left( \cos \left( \phi \right)
 \right) ^{-{\frac {p-2}{p}}} \left( \sin \left( \phi \right)
 \right) ^{-{\frac {-2+k}{k}}}{p}^{-1}{k}^{-1}\,d\phi \le $$ $$ \int_1^\infty r^{-{\frac {pk-2\,k-2\,p}{pk}}-2} \,dr\int_{0}^{\pi/2}\left( \cos \left( \phi \right)
 \right) ^{-{\frac {p-2}{p}}} \left( \sin \left( \phi \right)
 \right) ^{-{\frac {-2+k}{k}}}{p}^{-1}{k}^{-1}\,d\phi. $$
 The inequalities appear because we extend the region of the integration from the quadrant $\{(x,y):x \ge 1, y \ge 1 \} $ to $\{(r,\phi): r\ge 1, \phi \ge \phi_1(r)=\arcsin(1/r), \phi \le \phi_2(r)=\pi/2-\arcsin(1/r)\} \subset$ $ \{(r,\phi):r\ge 1,\, \phi \ge 0, \,\phi \le \pi/2\}.$ The integral $$ \int_0^{\pi/2}\left( \cos \left( \phi \right)
 \right) ^{-{\frac {p-2}{p}}} \left( \sin \left( \phi \right)
 \right) ^{-{\frac {-2+k}{k}}}{p}^{-1}{k}^{-1}\,d\phi$$ converges because $p>1,\,k>1.$ The integral $$ \int_1^\infty r^{-{\frac {pk-2\,k-2\,p}{pk}}-2} \,dr$$ converges iff $$\frac 1 p +\frac 1 k <1. $$
