Prove that the given series $\sum_{(m,n)∈\mathbb{Z^2})} |\max\{m,n\}|^{-5}$ is convergent.

Question

Show that the given series $S$ is convergent where $$S=\sum_{(m,n)∈\mathbb{Z^2}} |{\max}\{m,n\}|^{-5}.$$ Also $(m,n)$ cannot be the point $(0,0)$. I am totally confused with this summation. If without loss of generality we take $\max\{m,n\}=m$ then the summation reduces to $\sum_{m∈\mathbb{Z}} |m|^{-5}$ which is convergent. But I think this is not correct any way, as maximum of $m$ and $n$ runs through the set of integers. Please help me to solve this.

Answer 1

$$\sum_{(m,n)\in\Bbb Z^2, \\(m,n)\neq(0,0)} \big(\max{|m|,|n|}\big)^{-5}= $$

$$=\sum_{(m,n)\in\Bbb Z^2,\\ |m|\neq |n|} \big(\max{|m|,|n|}\big)^{-5}+\sum_{(m,n)\in\Bbb Z^2\\ |m|=|n|} |m|^{-5}$$

The series converges if and only if both summands converge. Now: $$\sum_{(m,n)\in\Bbb Z^2\\ |m|=|n|} |m|^{-5}=4\sum_{m\in \Bbb N} m^{-5}<4\sum_{m\in\Bbb N}m^{-2}<+\infty$$

Finally:

$$\sum_{(m,n)\in\Bbb Z^2,\\ |m|\neq |n|} \big(\max{|m|,|n|}\big)^{-5}=2\sum_{|m|>|n|}\big(\max{|m|,|n|}\big)^{-5}=2\cdot4\sum_{m>n>0}\big(\max{|m|,|n|}\big)^{-5}$$

$$2\cdot4\sum_{(m,n)\in\Bbb Z^2\\m>n>0} |m|^{-5}=2\cdot 4\sum_{m=1}^\infty\sum_{n=0}^{m-1} |m|^{-5}=8\sum_{m=1}^\infty\frac{(m-1)(m-2)}{2}\cdot\frac{1}{m^5}=$$ $$=4\sum_{m=1}^\infty\frac{m-1}{m}\frac{m-2}{m}\frac{1}{m^3}<4\sum_{m=1}^\infty\frac{1}{m^3}<\infty$$

Answer 2

In order to have convergence the series should be something like this (see Hayk's comment) $$\sum_{(m,n)\in\mathbb{Z}^2\\(m,n)\not=(0,0)} (\max\{|m|,|n|\})^{-d}$$ with $d>2$.

Since the terms of the series are positive we can rearrange them and rewrite the series as $$\begin{align*} \sum_{(m,n)\in\mathbb{Z}^2\\(m,n)\not=(0,0)} (\max\{|m|,|n|\})^{-d}&\leq 4 \sum_{(m,n)\in\mathbb{N}^2\\(m,n)\not=(0,0)} (\max\{m,n\})^{-d}\\ &= 4\sum_{k=1 }^{\infty}\sum_{m+n=k}(\max\{m,n\})^{-d}\\ &\leq 4\sum_{k=1 }^{\infty}\sum_{m+n=k} (k/2)^{-d}\\ &=2^{d+2}\sum_{k=1 }^{\infty} (k+1)\cdot k^{-d}\\&= 2^{d+2}\sum_{k=1 }^{\infty} \frac{1}{k^{d-1}}+2^{d+2}\sum_{k=1 }^{\infty} \frac{1}{k^d}<+\infty. \end{align*}$$