Convergence of $\sum_{n_{1}=1}^{\infty}\cdots\sum_{n_{k}=1}^{\infty}\frac{1}{\left(n_{1}^{2}+\dots+n_{k}^{2}\right)^{\alpha}}$

Question

Let $\alpha >0$ a real number and $k>0$ an integer. I wolud like to know for which $\alpha$ the multiple series $$\sum_{n_{1}=1}^{\infty}\cdots\sum_{n_{k}=1}^{\infty}\frac{1}{\left(n_{1}^{2}+\dots+n_{k}^{2}\right)^{\alpha}}$$ converges.

My try: I tried to study the following integrals $$\int_{1}^{+\infty}\cdots\int_{1}^{+\infty}\frac{1}{\left(x_{1}^{2}+\dots+x_{k}^{2}\right)^{\alpha}}dx_{1}\cdots dx_{k}$$ even if I'm not sure it is the right thing to do, since I have not found the n-dimentional integral test. Furthermore, analyzing only an integral and making a change of variable, I obtain $$\int_{1+x_{2}^{2}+\dots+x_{k}^{2}}^{+\infty}\frac{1}{v^{\alpha}}\frac{1}{2\sqrt{v-x_{2}^{2}-\dots-x_{k}^{2}}}dv$$ and now I'm stuck again.

Answer 1

Let $$ r_k(n) = \left|\left\{(a_1,\ldots,a_k)\in\mathbb{Z}^k:a_1^2+\ldots+a_k^2=n\right\}\right| $$ (which, for $k=2$, is given by $4(\chi_4*1)(n)$ since $\mathbb{Z}[i]$ is a Euclidean domain). This function counts the number of lattice points on a hypersphere. By a small variation of Gauss well-known argument, if we denote as $N_k(R)$ the number of lattice points in the region $a_1^2+\ldots+a_k^2\leq R^2$ we have $$ \sum_{n\leq R^2} r_k(n) =N_k(R) = \frac{\pi^{k/2}}{\Gamma\left(1+k/2\right)} R^{k} + O(R^{k-1}) $$ hence $$ \sum_{n\leq M}r_k(n) = \frac{\pi^{k/2}}{\Gamma(1+k/2)} M^{k/2} + O(M^{k/2-1/2}) $$ which is enough to discuss the convergence of your series by invoking summation by parts.
Indeed the given series behaves like a multiple of $$ \sum_{n\geq 1}\frac{r_k(n)}{n^{\alpha}} $$ which by summation by parts behaves like a multiple of $$ \sum_{n\geq 1}\frac{n^{k/2}}{n^{\alpha+1}}. $$ It follows that the series is convergent for $\color{red}{\alpha>\frac{k}{2}}$, just like the mentioned integral.

For $k=2$ and $\alpha>1$ we have

$$ \sum_{n,m\geq 1}\frac{1}{(n^2+m^2)^\alpha}=\frac{1}{4}\left(\sum_{n\geq 1}\frac{r_2(n)}{n^\alpha}-4\zeta(2\alpha)\right)=\beta(\alpha)\zeta(\alpha)-\zeta(2\alpha)$$ by Dirichlet's convolution and we also have reasonably simple closed formulas for $k=4$.

Answer 2

Several considerations.

As $$\frac{1}{\left(n_{1}^{2}+\dots+n_{k}^{2}\right)^{\alpha}} < \frac{1}{\left(n_{1}^{2}\right)^{\alpha}}$$ then for $2\alpha > 1$ we have converged case.

Now let's take $0 <2\alpha \leqslant 1$ and consider $\left(n_{1}^{2}+\dots+n_{k}^{2}\right)^{\alpha} \leqslant \left(n_{1}+\dots+n_{k}\right)^{2\alpha} $ which gives $$\frac{1}{\left(n_{1}^{2}+\dots+n_{k}^{2}\right)^{\alpha}} \geqslant \frac{1}{\left(n_{1}+\dots+n_{k}\right)^{2\alpha} }$$ If we take partial sums from 1 up to $N_1,N_2,\cdots,N_k$ then they always contain partial sum for corresponding harmonic type series from $k$ up to $N_1+N_2+\cdots+N_k$

Answer 3

A “faster than light” proof. By AM-GM inequality we have $$\sum_{n_{1}\geq1}\cdots\sum_{n_{k}\geq1}\frac{1}{\left(n_{1}^{2}+\dots+n_{k}^{2}\right)^{\alpha}}\leq\frac{1}{k^{\alpha}}\sum_{n_{1}\geq1}\frac{1}{n_{1}^{2\alpha/k}}\cdots\sum_{n_{k}\geq1}\frac{1}{n_{k}^{2\alpha/k}}.$$