$ \sum_{m=1}^\infty\sum_{n=1}^{\infty}\frac{1}{|m+in|^p}<\infty $

Question

I need some help with this problem:

$$ \text{If} \;p>2 \Longrightarrow \sum_{m=1}^\infty\sum_{n=1}^{\infty}\frac{1}{|m+in|^p}<\infty $$

Note that $$\sum_{m=1}^\infty\sum_{n=1}^{\infty}\frac{1}{|m+in|^2}=\infty$$

$|m+in|^2=m^2+n^2$

Thanks in advance.

Answer 1

We have $$\sum_{m,n=1}^{m^2+n^2\leq R^2} \dfrac1{(m^2+n^2)^{p/2}} < \iint_{x^2+y^2 \leq R^2} \dfrac{dxdy}{(x^2+y^2)^{p/2}} = \int_{r=0}^R\int_{t = 0}^{2 \pi} \dfrac{rdrdt}{r^p} = 2\pi \int_{r=0}^R \dfrac{dr}{r^{p-1}}$$ Now conclude what you want.

You can also show that $p>2$ is necessary by lower-bounding $$\sum_{m,n=1}^{m^2+n^2\leq R^2} \dfrac1{(m^2+n^2)^{p/2}} > \iint_{x^2+y^2 \geq 1}^{x^2+y^2 \leq R^2} \dfrac{dxdy}{(x^2+y^2)^{p/2}} = \int_{r=1}^R\int_{t = 0}^{2 \pi} \dfrac{rdrdt}{r^p} = 2\pi \int_{r=1}^R \dfrac{dr}{r^{p-1}}$$