TITLE: Absolute convergence of $\sum \frac {1} {\left( m_{1}^{2}+m_{2}^{2}+\cdots +m_{r }^{2}\right)^{\mu} } $
QUESTION [5 upvotes]: I am trying to prove that the series $\sum \dfrac {1} {\left( m_{1}^{2}+m_{2}^{2}+\cdots +m_{r }^{2}\right)^{\mu} } $ in which the summation extends over all positive and negative integral values and zero values of $m_1, m_2,\dots, m_r$, except the set of simultaneous zero values, is absolutely convergent if $\mu > \dfrac {r} {2}$.
Any help with a proof strategy would be much appreciated.

REPLY [3 votes]: Here is a way I like. We can rewrite your sum as $$\sum_{\boldsymbol{m}\in\mathbb{Z}^{m}\backslash\{\boldsymbol{0}\}}\frac{1}{\|\boldsymbol{m}\|_{2}^{r+\epsilon}}$$ where $\epsilon>0.$ Then since $$\|x\|_{2}\geq\max_{i}|x_{i}|,$$ by using the comparison test, we know that our original series will converge if  $$\sum_{\boldsymbol{m}\in\mathbb{Z}^{m}\backslash\{\boldsymbol{0}\}}\frac{1}{\max_{i}|\boldsymbol{m}_{i}|^{r+\epsilon}}$$ converges. Since the set of all $\boldsymbol{m}$  with $k-1\leq\max_{i}|\boldsymbol{m}_{i}|\leq k$  has size $\leq Ck^{r-1}$ for some constant $C,$ (it is the surface of an $r$  dimensional cube) we see that the above is bounded by $$\sum_{k=1}^{\infty}C\frac{k^{r-1}}{k^{r+\epsilon}}\leq C\sum_{k=1}^{\infty}\frac{1}{k^{1+\epsilon}}$$ which converges.