I would like a reference/proof for the following statement:

Statement.For $p\in\mathbb R$, the sum $$\sum_{\mathbf k\in\mathbb Z^n\setminus\{0\}} \lvert\mathbf k\rvert^{-p}$$

converges (in the sense of the Lebesgue integral with respect to the counting measure) if and only if $p>n$, where $\lvert\mathbf k\rvert$ is the usual Euclidean norm of a vector. (I.e. the square root of the sum of its coordinates.)

*Example.* For $n=1$, this is equivalent to saying that $\sum_{k=1}^\infty k^{-p}$ converges if and only if $p>1$, which is a classically well-known result.

The obvious way to proceed in proving the statement would be to compare the sum to the integral $$\int \lvert x\rvert^{-p} \, dx,$$ which can then be treated using polar coordinates. However, getting the integration bounds right is tedious. Therefore I was wondering if there is some solid reference for this.

## Best Answer

I just noticed that this is essentially reuns comment. But anyway...

The sum is of positive terms so allow the value $\infty$ and then we don't need to tip-toe around convergence issues$^A$. The divergence for $p\le 1$ is immediate as the 1D sum is smaller than this sum, being a sum over only $(0,\dots,0,k_n)\in\mathbb Z^n\setminus\{0\}$.

Since $k\in\mathbb Z^n\setminus\{ 0\}$, $|k|\ge 1$ and hence $2|k|^2\ge |k|^2 + 1.$ There's also an obvious upper bound $|k|^2 \le |k|^2+1$. Hence, up to an irrelevant constant, it suffices to consider the finiteness of $$\sum_{k\in\mathbb Z^d} (1+|k|^2)^{-p/2} $$

On the box of size 1 around $k$, $x\in\prod _i [k_i-1/2,k_i+1/2]=:C_k$, we have $$ x_i^2 \le (k_i + 1/2)^2 \le 2k_i^2 + 1/2, \implies |x|^2+1\le 2|k|^2 +n+1 \le (n+1)(|k|^2+1),$$ and since we also have $k\in\prod _i [x_i-1/2,x_i+1/2]$, $$ \frac1{n+1}(|x|^2+1)\le |k|^2 +1 \le (n+1)( |x|^2+1). $$

The boxes $C_k$ partition $\mathbb R^n$ (up to null sets$^B$) and have measure 1. Hence, integrating the above inequality over each $C_k$ and summing, we obtain (having dispensed of the case $p<0$) $$\frac1{(n+1)^{p/2}}\int_{\mathbb R^n} (|x|^2+1)^{-p/2} dx \le \sum_{k\in\mathbb Z^d} (1+|k|^2)^{-p/2} \le (n+1)^{p/2} \int_{\mathbb R^n} (|x|^2+1)^{-p/2} dx $$ and the result follows from the easy result for integrals$^C$.

$^A$: An approach I am confident in suggesting to someone who tries to understand a sum as an integral over counting measure...

$^B$: easy from first principles, as the intersections can be written as a countable union of hyperplanes, which are themselves null.

$^C$: I see that your own answer requires explicitly writing out the change of variables to polar. This isn't necessary for the stated result. By the AM-GM inequality, $$ ((1+x_1^2)(1+x_2^2)\dots (1+x_n^2))^{1/n} \le 1 +x_1^2 + \dots + x_n^2 . $$ and hence by Tonelli's theorem, the finiteness for $p>n$ follows from the 1D result. Up to a multiplicative constant depending only on $n$, the AM-GM inequality can also be reversed. So the divergence result for $p\le n$ holds as well.

...OK, I don't know an elementary proof of the reverse AM-GM, but you can also use a much simpler change of variables, namely scaling. I can't remember the formula for polar coordinates but I could probably drunk-prove $d(\lambda x) = \lambda^n dx$. The above establishes that it is enough to check the convergence/divergence of $$ \int_{|x|>1} |x|^{-p} dx = \sum_{k=0}^\infty \int_{2^k <|x|<2^{k+1}} |x|^{-p} dx = \left(\sum_{k=0}^\infty 2^{k(n-p)}\right) \int_{1<|x|<2}|x|^{-p} dx. $$ as $|x|^{-p} \in L^1( 1<|x|<2 )$, the LHS integral diverges or converges as $\sum_{k=0}^\infty 2^{k(n-p)}$ does, QED.