Summability of the Negative Powers of the Fourier Weights of the Torus

In the setting of The Real Numbers: Standing Notation and Background, let $n,s\in\mathbb{N}$ satisfy $n\le s$, and let $[n]$ be the initial segment determined by $n$. Let $\mathbb{Z}$ be the set of integers, let $\mathbb{R}^{n}$ be Euclidean space, whose points are read as maps on $[n]$ with real values, the component of $k\in\mathbb{R}^{n}$ at $i\in[n]$ being written $k_{i}$, let $\lVert\,\cdot\,\rVert$ be the Euclidean norm, and let $\mathbb{Z}^{n}$ be the integer lattice. Convergence of a series of real numbers and its sum are as defined there, finite products are those of Finite Product Notation in a Field, and $2=1+1$, $4=2+2$.

Let $\pi$ be the real number of The Number Pi (i). It is positive: $\pi=2x_{0}$ with $0<x_{0}$ by The Least Positive Zero of the Cosine (i), and $0<2$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, so $0<\pi$ by claim 5 of that lemma. Hence $\pi^{2}$ is positive by claim 5 again, and $\tfrac{1}{\pi^{2}}$ is defined by claim 7 of that lemma. For $k\in\mathbb{Z}^{n}$ put

$$\mu_{k}=1+4\pi^{2}\lVert k\rVert^{2},$$

and let $\mu_{k}^{s}$ be the natural power. Then the following hold.

1. (Positivity and a product bound)(i) For every $k\in\mathbb{Z}^{n}$ one has $1\le\mu_{k}$; consequently $\mu_{k}$ and $\mu_{k}^{s}$ are positive, their inverses are defined, and

$$\frac{1}{\mu_{k}^{s}}\le\prod_{i=1}^{n}\frac{1}{1+4\pi^{2}k_{i}^{2}},$$

each factor on the right being defined because its denominator is positive.

2. (Summability)(ii) Let $\kappa:\mathbb{N}\to\mathbb{Z}^{n}$ satisfy: $\kappa(j)=\kappa(j')$ implies $j=j'$. Then the series $\sum_{j=1}^{\infty}\tfrac{1}{\mu_{\kappa(j)}^{s}}$ converges, and

$$\sum_{j=1}^{\infty}\frac{1}{\mu_{\kappa(j)}^{s}}\le\Bigl(1+\frac{1}{\pi^{2}}\Bigr)^{n}.$$