I came across a problem concerning about the convergence of products. I wonder if the complex series $a\_n=\prod^n\_{k=1}(1-e^{k\alpha \pi i})$ converges to zero when $\alpha$ is irrational. Of course, the infinite product doesn't converge to any nonzero number.


It seems that the behavior of $e^{k\alpha \pi i}$ is not quite "predictable". I have no idea how to approach this problem. Thanks in advance!



Does $a_n=\prod^n_{k=1}(1-e^{k\alpha \pi i})$ converge to zero when $\alpha$ is irrational?
The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that
$$ 
\limsup\_{N\to \infty} \prod\_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1}
$$
(Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$).


I'll prove a slightly weaker result; namely I'll show that (1) holds for any irrational $\alpha$ for which there are infinitely many rational approximations $a/q$ with $(a,q)=1$ and
$$ 
\alpha=\frac{a}{q} +\beta, \qquad \text{with}\qquad |\beta| \le \frac{1}{100q^2}. \tag{2}
$$
The $1/100$ is just chosen for ease of exposition, and a more careful argument can make do with $1/\sqrt{5}$ (any constant $<1/2$ is enough), and then it is well known that every irrational number admits infinitely many approximations with $|\alpha -a/q| \le 1/(\sqrt{5}q^2)$. Condition 2 holds for almost all irrational numbers -- it fails only if the continued fraction expansion only uses numbers below $100$.


Suppose then that $q$ is such that (2) holds, and consider the product in question at $N=q-1$. By the triangle inequality, for $1\le n\le N$,
$$ 
|1-e(n\alpha)| = |(1-e(an/q))+e(an/q)(1-e(\beta n))| \ge |1-e(an/q)| 
\Big(1-\frac{|1-e(n\beta)|}{|1-e(an/q)|}\Big)= 2\Big|\sin\frac{\pi an}{q}\Big| \Big( 1- \frac{|\sin(\pi n\beta)|}{|\sin(\pi an/q)|}\Big).
$$
Now write $\Vert x\Vert= \min\_{\ell \in {\Bbb Z}} |x-\ell|$ for the distance from $x$ to the nearest integer. Note also that for $0\le x\le \pi/2$ one has $2x/\pi \le \sin x \le x$. Therefore we get for $1\le n\le N$
$$ 
|1-e(n\alpha)| \ge |1-e(an/q)| \Big(1 - \frac{\pi n|\beta|}{2\Vert an/q\Vert}\Big) \ge |1-e(an/q)| \exp\Big( - \frac{\pi q|\beta|}{\Vert an/q\Vert}\Big), \tag{3}
$$
where in the last inequality we used that $\eta =\pi n|\beta|/(2\Vert an/q\Vert) \le \pi q^2|\beta|/2 \le 1/10$, so that $(1-\eta) \ge \exp(-2\eta)$.


Multiplying (3) for $n$ from $1$ to $N=q-1$, we obtain
$$ 
\prod\_{n=1}^{N} |1-e(n\alpha)| \ge \prod\_{n=1}^{q-1} |1-e(an/q)| 
\exp\Big( - \sum\_{n=1}^{N} \frac{\pi q |\beta|}{\Vert an/q\Vert}\Big). \tag{4}
$$
Now
$$ 
\sum\_{n=1}^{q-1} \frac{1}{\Vert an/q\Vert} \le 2\sum\_{n\le q/2} \frac{q}{n} =2 q \log q +O(q),
$$
and $\prod\_{n=1}^{N} |1-e(an/q)| = q$, and so from (4) we conclude that
$$ 
\prod\_{n=1}^{q-1} |1-e(n\alpha)| \ge q \exp\Big( - 2\pi q^2 |\beta| \log q +O(q^2|\beta|)\Big) \gg q^{0.9}. 
$$


This proves the claim. Note that this argument is closely related to the nice attempt of Sangchul Lee.


**Added information (October, 2021):** From comments on Fedja's recent question [Can all partial sums $\sum\_{k=1}^n f(ka)$ where $f(x)=\log|2\sin(x/2)|$ be non-negative?](https://mathoverflow.net/questions/406249/can-all-partial-sums-sum-k-1n-fka-where-fx-log2-sinx-2-be-non-n) I was led to the [following paper by D.S. Lubinsky](https://www.sciencedirect.com/science/article/pii/S0022314X98923654) which shows that (see Theorem 1.2 there)
$$
\limsup\_{N\to \infty} \frac{\log \prod\_{n=1}^{N} |1-e(n\alpha)|}{\log N} \ge 1,
$$
for all irrational numbers $\alpha$, so that the product in the question gets almost as large as $N$ infinitely often.


