Convergence of double sum $\sum_{m, n}\frac{1}{m^p + n^k}$

Question

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.

Answer 1

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 $\frac1p+\frac1k\lt1$.

Answer 2

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 $$\sum_{m=1}^\infty m^{-p(1-\frac{1}{k})}<\infty \iff p(1-\frac{1}{k})>1.$$

Answer 3

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.

Answer 4

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 $$\frac 1 p +\frac 1 k <1. $$