$\sum_{\omega\in \mathbb{Z}(i)^*} |\omega|^{-2}$ does not converge.

Question

I'm trying to prove that the series $$ \sum_{\omega\in \mathbb{Z}(i)^*} |\omega|^{-2} $$

The problem can be viewed as the sum above the fundamental region $\Omega^* = \{m\omega_1 + n\omega_2 : m,n\in \mathbb{z}\}\backslash\{0\}$ for $\omega_1 = 1$ and $\omega_2 = i$. Note: $\mathbb{Z}(i) = \{m + ni : m,n\in\mathbb{Z}\}$ and $\mathbb{Z}(i)^*=\mathbb{Z}(i)\backslash\{0\}$.

Any hint? I thought it will suffice to use that for $m\geq n > 0$ we have $|m+ni|\leq 2m$, but I get nothing from that... Thank you.

Answer 1

First, since we're in a finite dimensional case, $|w|$ and $\|w\|_\infty$ (the latter defined as $\|a+bi\|_\infty=\max(|a|,|b|)$ are equivalent norms. Therefore $$\sum |w|^{-2}\geq K\sum\|w\|_\infty^{-2}$$ for some constant $K$, and it is enough to show that the latter series diverges.

Now by an easy induction there are exactly $8k$ elements $w$ such that $\|w\|_\infty=k$, for any positive $k$. Therefore: $$\sum\|w\|_\infty^{-2}=\sum_{k=1}^\infty \frac{8k}{k^2}$$ which clearly diverges.

Answer 2

Hint: when $R > 0$ is large, there are about $R^2$ values of $\omega \in \mathbb Z[i]^*$ for which $R \leq |\omega| \leq 2R$.

Answer 3

Presumably this is $$\sum_{a,b}\frac1{a^2+b^2}$$ is divergent, where the sum is over all integers $a$, $b$ with $(a,b)\ne(0,0)$. I'll show that the corresponding sum where we assume $a\ge0$, $b\ge0$, $(a,b)\ne(0,0)$ is divergent.

Note that $$\frac1{a^2+b^2}\ge\int_a^{a+1}\int_b^{b+1}\frac{dx\,dy}{x^2+y^2}$$ and so $$\sum_{a\ge0,b\ge0,(a,b)\ne(0,0)}\frac1{a^2+b^2} \ge\int\!\!\!\int_S\frac{dx\,dy}{x^2+y^2} \ge\int\!\!\!\int_T\frac{dx\,dy}{x^2+y^2}$$ where $S$ is the positive real quadrant minus the square $[0,1]^2$ and $T$ is the positive real quadrant minus the quarter-disc with $x^2+y^2\le 2$. But switching to polars gives $$\int\!\!\!\int_T\frac{dx\,dy}{x^2+y^2}=\infty.$$

Answer 4

One may prove that $\sum_{l \in L\ \{0\}} \frac{1}{|l|^a}$ converges iff a>2, where L is a lattice.

I will provide a proof for the ( $\rightarrow$). Let $b_1,b_2$ be the generators of the lattice, and $m,n$ be any integers. Then:

$|mb_1+nb_2|=|b_1||m+n\frac{b_2}{b_1}| \geq |b_1||n||Im(\frac{b_2}{b_1})|$

Similarly, $|mb_1+nb_2| \geq |b_2||m||Im(\frac{b1}{b2})|$, and put $k=min(|b_1 ||Im(\frac{b_2}{b_1})|, |b_2||Im(\frac{b1}{b2})|)$

Then $|mb_1+nb_2| \geq k|n|$ and $|mb_1+nb_2| \geq k|m|$ which implies that $|mb_1+nb_2| \geq k.max(|n|,|m|)$

Then we have: $\sum_{l \in L\ \{0\}} \frac{1}{|l|^a} \leq \sum_{(m,n) \neq 0} \frac{1}{k^amax(|n|,|m|)^a} \leq \frac{1}{k^a}\sum_{u \neq 0} \frac{8u}{u^a}$, where u=max(|n|,|m|).

This converges if $a-1 \geq 1$

Answer 5

This can be tackled through the elementary bounds for the Gauss circle problem. By introducing $$ N(R) = \left|\{(a,b)\in\mathbb{Z}^2: a^2+b^2\leq R^2\}\right| $$ we have $N(R)=\pi R^2 + O(R)$ by square-counting arguments. Let $$ R(n) = \left|\{(a,b)\in\mathbb{Z}^2 : a^2+b^2 = n\}\right|. $$ We have $\sum_{n=1}^{N}R(n) = N(\sqrt{n}) = \pi n + O(\sqrt{n})$, then

$$ \sum_{\substack{(a,b)\neq (0,0)\\ a^2+b^2\leq T^2}}\frac{1}{a^2+b^2}=\sum_{n=1}^{T^2}\frac{R(n)}{n}=\frac{N(T)}{T^2}+\sum_{n=1}^{T^2-1}\frac{N(\sqrt{n})}{n(n+1)} $$ by summation by parts. On the other hand $\frac{N(T)}{T^2}=\pi+o(1)$ and $$ \sum_{n=1}^{T^2-1}\frac{N(\sqrt{n})}{n(n+1)} = \pi\sum_{n=1}^{T^2-1}\frac{1}{n+1}+O(1) = 2\pi\log T+O(1), $$ hence the given series is divergent.