Absolute convergence of $\sum \frac {1} {\left( m_{1}^{2}+m_{2}^{2}+\cdots +m_{r }^{2}\right)^{\mu} } $

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.

Comparing the series with an integral, one sees that the series converges if and only if the $r$-dimensional integral $$ I_r=\int_{\mathbb R^r}[\|x\|\geqslant1]\,\frac{\mathrm dx}{\|x\|^{2\mu}} $$ converges. Consider the spherical coordinates $(s,\alpha)$ with $s\geqslant0$ and $\alpha$ in the sphere $S^{r-1}$. Then $\mathrm dx$ is proportional to $s^{r-1}\mathrm ds\mathrm d\alpha$. Hence , $I_r$ converges if and only if the $1$-dimensional integral $$ \int_1^{+\infty}\frac{s^{r-1}\mathrm ds}{s^{2\mu}} $$ converges, that is, if and only if $2\mu\gt r$.