Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to prove that the series $\sum \dfrac {1} {\left( m_{1}^{2}+m_{2}^{2}+\ldots +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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

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 $2\mu\gt r$.

share|cite|improve this answer

What about comparing with an integral on $\mathbb{R}^r$? And then an appropriate change of variable?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.