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

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.

• No idea what is unclear to you. It is obvious that $|k|^{-p}\sim \int_{k+[0,1]^n} |x|^{-p}dx$ so $\sum_{k\ne 0} |k|^{-p}$ converges iff $\sum_{|k|>1} \int_{k+[0,1]^n} |x|^{-p}dx$ converges iff $\int_{|x|>1 } |x|^{-p}dx$ converges Commented Jan 10, 2022 at 10:21
• @reuns You are totally right, but I was trying to change a lightbulb (see my way too over-formal answer below ðŸ™‚). Commented Jan 10, 2022 at 11:27

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.

• Thanks! FYI I only asked this question because I used the result here and the argument that I originally gave was formally wrong :) Commented Jan 10, 2022 at 14:31
• @MaximilianJanisch you're welcome; I don't at a glance see the result being used, but hope it is helpful nonetheless Commented Jan 10, 2022 at 14:47
• Would appreciate feedback for the downvote! Commented Jan 11, 2022 at 6:23

Another approach that avoids using polar coordinates to evaluate an integral is to use "polar coordinates" for the sum. It suffices to consider the sum $$\sum_{k \in \mathbb{N}^n}|k|^{-p}.$$ Since all norms on $$\mathbb{R}^n$$ are equivalent, we can use $$|k| = \max(|k_1|, \dots, |k_n|)$$. For $$r \in \mathbb{N}$$, let $$F(r) = \#\{k \in \mathbb{N}^n : |k| \leq r\} = r^n,$$ $$f(r) = \#\{k \in \mathbb{N}^n : |k| = r\} = F(r) - F(r - 1) = r^n - (r - 1)^n.$$ Then $$\sum_{k \in \mathbb{N}^n}|k|^{-p} = \sum_{r = 1}^{\infty}f(r)r^{-p} = \sum_{r = 1}^{\infty}r^{-p}(r^{n} - (r - 1)^n).$$ Note $$r^n - (r - 1)^n \sim nr^{n - 1}$$ as $$r \to \infty$$. Hence $$r^{-p}(r^{n} - (r - 1)^n) \sim nr^{-p + n - 1}$$. Applying the $$n = 1$$ case gives the result.

• Just one observation, minor, $F(r) = (r+1)^n$, but changes nothing. +1, neat. Commented Jan 11, 2022 at 1:24
• @orangeskid I'm using $\mathbb{N} = \{1, 2, \dots\}$. So $|k| \leq r$ if and only if $1 \leq k_j \leq r$ for each $j$. So wouldn't there be $r^n$ possible values of $k$ with $|k| \leq r$? Commented Jan 11, 2022 at 2:57
• Oh, I see, then your numbers are correct; the lattice points on the coordinate plane are possibly left out, although it does not matter in the final result. Still much neater than the convoluted approaches with integrals. Cheers! Commented Jan 11, 2022 at 3:23
• already +1'd, but disagree that the integrals are convoluted :) the only reason I bothered to type anything longer than reuns' comment is due to the length of the OP's own answer, and hence strived for something along the same lines, from first principles but simpler. Commented Jan 11, 2022 at 6:48
• @CalvinKhor The integral approach is the first that came to my mind, and it is quite simple if you know the general formula for the Lebesgue integral in polar coordinates, which follows easily from computing some metric tensors. But this approach is a way to circumvent using it. To prove it using comparison with the integral, for say, $n = 2$, for each square on the plane, you upper bound $|x|$ by the norm $|k|$ of the upper right corner of the square, and similarly you lower bound $|x|$ by the lower left corner of the square. This is complteley analgous to case $n = 1$. Commented Jan 11, 2022 at 6:57

In this answer, I will make my intuition of comparing the series to an integral rigorous.

From now on, $$n$$ always denotes any natural number and a vector $$x\in\mathbb R^n$$ or $$x\in\mathbb Z^n$$ will be written as $$x=(x^1,\dots, x^n)$$.

Theorem. 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 $$\lvert\mathbf k\rvert=\sqrt{\sum_{j=1}^n (\mathbf k^j)^2}$$ denotes the usual Euclidean norm of $$\mathbf k$$.

Lemma 1. For any $$n\in\mathbb N$$ and $$p\in\mathbb R$$, $$\sum_{\mathbf k\in\mathbb Z^n\setminus\{0\}} \lvert\mathbf k\rvert^{-p}=2^n\sum_{\mathbf k\in\mathbb N^n}\lvert\mathbf k\rvert^{-p},$$ where both sides may be infinite.

Lemma 2. For $$p\in\mathbb R$$, the integral $$\int_{[1,\infty[^n} \lvert x\rvert^{-p}\,\mathrm dx,$$ in the Lebesgue sense, is finite if and only if $$p>n$$.

Lemma 3. For any $$p\in\mathbb R$$, $$\sum_{\mathbb k\in(\mathbb N\setminus\{1\})^n}\lvert\mathbf k\rvert^{-p}\le\int_{[1,\infty[^n} \lvert x\rvert^{-p}\,\mathrm dx.$$

Proof of the part of the Theorem where $$p>n$$ (by strong induction over $$n$$). By Lemma 1, the statement to prove is equivalent to $$\sum_{\mathbf k\in\mathbb N^{n}}\lvert\mathbf k\rvert^{-p}<\infty.$$

Base case ($$n=1$$). Follows directly from improper Riemann integration and the monotone convergence Theorem.

Induction step. Fix $$n\in\mathbb N$$ and suppose that $$\sum_{\mathbf k\in\mathbb N^{m}}\lvert\mathbf k\rvert^{-p}<\infty$$ for all $$m\in\mathbb N\cap[1,n-1]$$. Note $$\mathbb N^n =\bigcup_{X\subset\{1,\dots,n\}} \mathbb N_X,$$ where $$\mathbb N_X\overset{\text{Def.}}=\{\mathbf k\in\mathbb N^n:\mathbf k^x=1\text{ for all }x\in X\text{ and }\mathbf k^x\neq 1\text{ for all }x\not\in X\}.$$

By Lemma 2 and 3, if $$p>n$$, then $$\sum_{\mathbf k\in\mathbb N_{\emptyset}} \lvert\mathbf k\rvert^{-p}<\infty$$. By the induction hypothesis, for any $$X\subset\{1,\dots,n\}$$ which is not the empty set, $$\sum_{\mathbf k\in\mathbb N_{X}} \lvert\mathbf k\rvert^{-p}<\infty.$$

This proves that $$\sum_{\mathbf k\in\mathbb N^{n}}\lvert\mathbf k\rvert^{-p}<\infty.$$ $$\square$$

Sketch of the proof of the part of the Theorem where $$p\le n$$. Use the same argumentation as in Lemma 3 to show that $$\sum_{\mathbf k\in\mathbb N^n}\lvert\mathbf k\rvert^{-p}\ge\int_{[1,\infty[^n}\lvert x\rvert^{-p}\,\mathrm dx$$ and conclude using the same argumentation as in Lemma 2.

Proof of Lemmas

Proof of Lemma 1. We may write $$\mathbb Z^n\setminus\{0\}=\bigcup_{s=(s_1,\dots,s_n)\in\{-1,1\}^n} \underbrace{\prod_{j=1}^n (-1)^{s_j} \mathbb N}_{\overset{\text{Def.}}=A_s},$$

where $$\prod$$ denotes the generalized cartesian product. Now we note that, since $$\lvert x\rvert$$ is invariant under the changing of sign of any coordinate of $$x$$, $$\sum_{\mathbf k\in A_{s}}\lvert\mathbf k\rvert^{-p}=\sum_{\mathbf k\in A_{\tilde s}} \lvert \mathbf k\rvert^{-p}$$ for any $$s,\tilde s\in \{-1,1\}^n$$. (Formally, this can be justified using the Transformationssatz (cf. for instance [1; Theorem 1]) for the pushforward under the flipping of coordinates of $$\mathbf k$$.)

The right-hand side of the Lemma is therefore $$\sum_{s\in\{-1,1\}^n}\sum_{\mathbf k\in A_{s}}\lvert\mathbf k\rvert^{-p},$$ which therefore equals the left-hand side from linearity of the sum. $$\square$$

Proof of Lemma 2. We use once again the Transformationssatz. Consider the function $$\begin{equation*}\begin{split}\Phi:]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[&\to\mathbb R^n\\\begin{pmatrix}r\\\phi_1\\\phi_2\\\dots\\\phi_{n-2}\\\phi_{n-1}\end{pmatrix}&\mapsto\begin{pmatrix}r\cos(\phi_1)\\r\sin(\phi_1)\cos(\phi_2)\\r\sin(\phi_1)\sin(\phi_2)\cos(\phi_3)\\\dots\\r\sin(\phi_1)\dots\sin(\phi_{n-2})\cos(\phi_{n-1})\\r\sin(\phi_1)\dots\sin(\phi_{n-2})\sin(\phi_{n-1})\end{pmatrix}.\end{split}\end{equation*}$$

Exercise. Prove that there exists a Lebesgue-null set $$N\subset\mathbb R^n$$ such that

1. $$\Phi:]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[\to\mathbb R^n\setminus N$$ is a $$C^1$$-diffeomorphism;
2. for all $$(r,\phi_1,\dots,\phi_{n-1})\subset]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[$$, we have $$\left\lvert\operatorname{det}\operatorname{Jac}\Phi((]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[)^\top)\right\rvert=r^{n-1}\sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\dots\sin(\phi_{n-2}).$$
3. $$\Phi(]1,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[)\supset[1,\infty[^n$$.

Hint. First, prove that $$\Phi$$ is injective. Second, note that, for a fixed $$r$$, the image of $$\Phi$$ is $$r\mathbb S^{n-1}$$, up to a Lebesgue-measurable set of Hausdorff dimension $$n-2$$. Then conclude that $$\Phi$$ is surjective up to a Lebesgue-measurable set of Hausdorff dimension $$n-1$$, i.e. of $$n$$-Lebesgue measure $$0$$. For the computation of the Jacobian, see https://en.wikipedia.org/wiki/N-sphere#Spherical_volume_and_area_elements.

Back to the proof of Lemma 2: From 1. and 3. above together with the Transformationssatz [1; Theorem 1] we get $$\int_{[1,\infty[^n} \lvert x\rvert^{-p}\,\mathrm dx\le2\pi^{n-1}\int_1^\infty r^{n-p-1}\,\mathrm dr.$$ This is Lebesgue-integrable iff $$p>n$$ by improper Riemann integration and the dominated convergence Theorem. $$\square$$

Proof of Lemma 3. This follows directly from $$\sigma$$-sub-additivity of the integral together with $$\int_{[\mathbf k^1-1,\mathbf k^1[\times\dots\times[\mathbf k^n-1,\mathbf k^n[}\lvert x\rvert^{-p}\ge\lvert\mathbf k\rvert^{-p}$$ for all $$\mathbf k\in\mathbb N^{n}$$.

[1]: Ivan Netuka, The Change-of-Variables Theorem for the Lebesgue Integral. ACTA UNIVERSITATIS MATTHIAE BELII, series MATHEMATICS 19 (2011), 37â€“42. Available online at https://actamath.savbb.sk/pdf/acta1906.pdf.

• Just a comment that you don't need the spherical coordinates transformation to prove your lemma 2. You can use the fairly elementary formula $$\int_{\mathbb{R}^n}f(x)\,dx = \int_{0}^{\infty}\int_{S^{n - 1}}f(r\omega)\,dS(\omega)r^{n - 1}\,dr$$ for measurable $f \geq 0$ to get $$\int_{|x| \geq 1}|x|^{-p}\,dx = \text{Area}(S^{n - 1})\int_{1}^{\infty}r^{-p + n - 1}\,dr.$$ Commented Jan 10, 2022 at 19:33
• Why so complicated? Commented Jan 11, 2022 at 3:31
• @orangeskid I wanted to understand some properties of the explicit $n$-sphere parametrization ðŸ™‚. Commented Jan 11, 2022 at 9:41
• @Maximilian Janisch: Fair enough. Btw, have you taken a look at Whittaker and Watson's book? Commented Jan 11, 2022 at 11:14
• @orange You mean this one ? I never read itâ€¦ Commented Jan 11, 2022 at 11:31