Summability of $\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\frac{1}{||k||^2}$

Question

It is well-known that the sum $\sum_{k\in\mathbb{Z}\setminus\{0\}}\frac{1}{k^2}$ exists.

Is it true that the sum$\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\frac{1}{||k||^2}$ exists? Here, $$||k||=\max_{1\leq j\leq d}|k_j|.$$ By partitioning the set $\mathbb{Z}^d$, it appears that this should diverge. What is the correct answer?

Answer 1

For a given $n > 0$ we have at least $ 2 \cdot d(2n)^{d-1}$ vectors such that :

$$\mid \mid k \mid \mid = n$$

Hence we have :

$$ S \geq \sum_{n \in \mathbb{N}^*} \frac{2d(2n)^{d-1}}{n^2}$$

Hence if $d \geq 2$ it diverges since the harmonic serie diverges.

Answer 2

For a positive integer $n$, the number of points $x$ in $\mathbb{Z}^d$ with $\|x\|=n$ is $$(2n+1)^d-(2n-1)^d=2^{d}dn^{d-1}+\mathcal{O}(n^{d-2})\,.$$ Thus, $$\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\,\dfrac{1}{\|k\|^2}=\sum_{n=1}^\infty\,\sum_{\substack{x\in\mathbb{Z}^d}\\{\|x\|=n}}\,\frac{1}{n^2}=\sum_{n=1}^\infty\,\frac{(2n+1)^d-(2n-1)^d}{n^2}\,.$$ Since $\sum\limits_{n=1}^\infty\,\dfrac{n^{d-1}}{n^2}$ diverges for all $d\geq2$, by the Comparison Test, $\sum\limits_{k\in\mathbb{Z}^d\setminus\{0\}}\,\dfrac{1}{\|k\|^2}$ diverges for all $d\geq 2$.

However, this sum converges for every $z\in\mathbb{C}$ with $\text{Re}(z)>0$: $$f_d(z):=\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\,\dfrac{1}{\|k\|^{d+z}}\,.$$ That is, $$f_d(z)=\sum_{r=1}^{\left\lfloor\frac{d}{2}\right\rfloor}\,2^{d-2r+2}\,\binom{d}{2r-1}\,\zeta(z+2r-1)\,,$$ where $\zeta$ is the Riemann zeta-function. In particular, $f_d(1)$ is a $\mathbb{Z}_{>0}$-linear combination of $\zeta(2)$, $\zeta(4)$, $\ldots$, $\zeta\Big(2\left\lfloor\frac{d}{2}\right\rfloor\Big)$: $$\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\,\dfrac{1}{\|k\|^{d+1}}=f_d(1)=\sum_{r=1}^{\left\lfloor\frac{d}{2}\right\rfloor}\,2^{d-2r+2}\,\binom{d}{2r-1}\,\zeta(2r)\,.$$

Answer 3

Divergence of the Euclidean Norm $$ \begin{align} \sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{j^2+k^2} &\ge\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{(j+k)^2-(j+k)}\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\left(\frac1{j+k-1}-\frac1{j+k}\right)\\ &=\sum_{j=1}^\infty\frac1j\tag1 \end{align} $$ which diverges.

$(1)$ is a part of the sum over $\mathbb{Z}^d$ for all $d\ge2$, so the sum diverges for all $d\ge2$.

---

Equivalent Norms

After I posted, I noticed that the norm in the question was not the Euclidean norm. However, $$ |k|^2=\sum_{j=1}^dk_j^2\ge\max_{1\le j\le d}(k_j)^2=\|k\|^2 $$ and $$ |k|^2=\sum_{j=1}^dk_j^2\le d\max_{1\le j\le d}(k_j)^2=d\|k\|^2 $$ Therefore, $$ \sum_{k\in\mathbb{Z}^d\setminus\{0\}}\frac1{|k|^2}\le\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\frac1{\|k\|^2}\le d\sum_{k\in\mathbb{Z}^d\setminus\{0\}}\frac1{|k|^2} $$ So the sum in the question diverges or converges the same as the sum with the Euclidean norm.