# Reference for convergence of $\sum_{\mathbf k\in\mathbb Z^3\setminus\{0\}} \lvert\mathbf k\rvert^{-p}$

convergence-divergencereal-analysisreference-requestsequences-and-series

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.

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

1. 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 $$(0,\dots,0,k_n)\in\mathbb Z^n\setminus\{0\}$$.

2. 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 $$\sum_{k\in\mathbb Z^d} (1+|k|^2)^{-p/2}$$

3. 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 $$k\in\prod _i [x_i-1/2,x_i+1/2]$$, $$\frac1{n+1}(|x|^2+1)\le |k|^2 +1 \le (n+1)( |x|^2+1).$$

4. 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.