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Let $s\in\mathbb R$. How to show the series $$\sum_{\xi\in\mathbb Z^n}\frac{1}{(1+|\xi|^2)^{s/2}}$$ converges if and only if $s>n$ ($n$ is the dimension of $\mathbb Z^n$)? The convergence of this series is to be understood as the existence of the limit $$\lim_{k\to \infty}\sum_{|\xi|\leq k}\frac{1}{(1+|\xi|^2)^{s/2}}.$$ Above $|\xi|$ indicates the usual Euclidian norm.

PtF
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2 Answers2

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Using standard methods, we get $$ \begin{align} \int_{\mathbb{R}^n}\frac{\mathrm{d}x}{\left(1+|x|^2\right)^{s/2}} &=n\omega_n\int_0^\infty\frac{r^{n-1}\,\mathrm{d}r}{\left(1+r^2\right)^{s/2}}\\ &=\frac{n}{2}\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}\int_0^\infty\frac{r^{n/2-1}\,\mathrm{d}r}{\left(1+r\right)^{s/2}}\\ &=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)}\frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{s-n}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\\ &=\pi^{n/2}\frac{\Gamma\left(\frac{s-n}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\tag{1} \end{align} $$ The integral in $(1)$ is decreasing in $s$, and as $s$ decreases to $n$, the integral grows like $$ \frac{c_n}{s-n}\tag{2} $$ where $c_n=\dfrac{2\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)}$. Thus, the integral in $(1)$ converges precisely when $s\gt n$.

Each point in the lattice sits at the center of its own unit cube with sides parallel to the axes. For the cube centered on the lattice point $k\in\mathbb{Z}^n$, when $|k|^2\ge n$, $$ \frac14\le\frac{1+(|k|-\sqrt{n}/2)^2}{1+|k|^2}\le\frac{1+|x|^2}{1+|k|^2}\le\frac{1+(|k|+\sqrt{n}/2)^2}{1+|k|^2}\le\frac94\tag{3} $$ Thus, $$ \sum_{k\in\mathbb{Z}^n}\frac1{\left(1+|k|^2\right)^s}\tag{4} $$ converges precisely when the integral in $(1)$ converges; that is, when $s\gt n$.

robjohn
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Here is not an answer, just some hints.

First way. Try to compare the original sum with the integral $$ \int\limits_{x\in\mathbb{R}^n}\frac{dx_1\cdots dx_n}{(1+\|x\|^2)^{s/2}}. $$ Start with $\int_{t\in\mathbb{R}}\frac{dt}{(a^2+t^2)^{s/2}}$, make a suitable change of variables and reduce it to the Beta function.

Second way. It is something like a "polar change of variables". Try to estimate the inner sum and reduce the original problem to a one-dimensional problem: $$ \sum_{p=0}^{\infty}\sum_{10p\le|\xi|<10p+10}\frac{1}{(1+\|\xi\|^2)^{s/2}}. $$

Egor Maximenko
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  • As you noticed, \begin{align} \displaystyle \sum_{\xi\in\mathbb Z^n} \frac{1}{(1+|\xi|^2)^s}&=\lim_{k\to \infty} \sum_{|\xi|\leq k} \frac{1}{(1+|\xi|^2)^s}\ &=\lim_{k\to \infty} \sum_{p=0}^{k} \sum_{p\leq |\xi|< p+1}\frac{1}{(1+|\xi|^2)^s}. \end{align} If $p\leq |\xi|<p+1$ then, \begin{align} \displaystyle (1+p^2)^{s/2}\leq (1+|\xi|^2)^{s/2} \end{align} – PtF Feb 09 '14 at 20:40
  • Consequently, \begin{align} \displaystyle \sum_{p\leq |\xi|<p+1}\frac{1}{(1+|\xi|^2)^{s/2}}\leq \sum_{p\leq |\xi|<p+1}\frac{1}{(1+p^2)^{s/2}}=\frac{C_p}{(1+p^2)^{s/2}}, \end{align} where $C_p$ is a constant that grows with $p$. So I must show $1/(1+p^2)^{s/2}$ decreases faster then $C_p$ grows when $p\to \infty$. Now I don't know how to do that, any idea? – PtF Feb 09 '14 at 20:40
  • Above I was reasoning firstly about the convergence of the series.. – PtF Feb 09 '14 at 20:53
  • @PtF, see simple estimate for the number of lattice points in a spherical shell at http://math.stackexchange.com/questions/670010/how-many-points-xi-in-mathbb-zn-are-there-satisfying-p-leq-xi-p1/670046#670046 – Will Jagy Feb 09 '14 at 22:05
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    Associate to every point $\xi=(\xi_1,\ldots,\xi_n)\in\mathbb{Z}^n$ the cube $K_\xi=[\xi_1,\xi_1+1)\times\cdots\times[\xi_n,\xi_n+1)$. If $10p\le |\xi|<10(p+1)$, then the cube $K_\xi$ is contained in ${x\in\mathbb{R}^n\colon 10p\le|x|<10(p+2)}$. It is a possible way to estimate (from above) the number of such points $\xi$. I put $10p$ in order to simplify also the lower estimate. But the problem purposed by Will Jagy is very interesting by itself. – Egor Maximenko Feb 10 '14 at 00:54
  • I am sorry, something like $10np$ or $10\sqrt{n}p$ should be used instead of $10p$. – Egor Maximenko Feb 10 '14 at 01:02