How to prove $\liminf_{n \to \infty} \prod_{k=1}^n |1-\exp(2\pi k i\alpha)|=0$ I wonder how to show the following limit inferior ($\alpha$ is an irrational number)
$$\liminf_{n \to \infty} \prod_{k=1}^n |1-\exp(2\pi k i\alpha)|=0.$$
Intuitively this is quite true since $\{\exp(2\pi k i\alpha)\}$ is uniformly distributed on the unit circle at origin (since elementary proofs are known, please feel free to use this result). When $n$ is large, $|1-\exp(2\pi k i\alpha)|$ is at most $2$ but can frequently be close to zero. So the product will be frequently close to zero (MatLab experiment also shows the same phenomenon). But how do we make this into a rigorous argument?
Another result we can consider using is the Huiwitz's theorem, which states that there are infinitely many relatively prime integers $a, q$ such that
$$|\alpha-\frac{a}{q}|&lt;\frac{1}{\sqrt{5}q^2}.$$
Let $(a_n),(q_n)$ be such a sequence with $q_n \to \infty$ and $\alpha=\frac{a_n}{q_n}+\beta$ and $|\beta|&lt;\frac{1}{\sqrt{5}q^2}.$ I speculate that
$$\lim_{n \to \infty} \prod_{k=1}^{q_n} |1-\exp(2\pi k i\alpha)|=0$$
(of course you don't have to use my idea to answer the question I asked)
Here is the reasoning for the result of limsup, which might help.
Let $a_n = \prod_{k=1}^n|1 - \exp(2\pi i \alpha k)|$
So
$$ a_n = \prod_{k=1}^n\sqrt{(\cos(2\pi\alpha k) - 1)^2 + \sin^2(2\pi\alpha k)}$$
$$ a_n = \prod_{k=1}^n\sqrt{2 - 2\cos^2(2\pi\alpha k)} $$
$$ a_n = 2^{\frac n2} \prod_{k=1}^n|\sin(2\pi\alpha k)|$$
Does this lead anywhere?
Let $a_n = \prod_{k=1}^n|1 - \exp(2\pi i \alpha k)|$
So
$$ a_n = \prod_{k=1}^n\sqrt{(\cos(2\pi\alpha k) - 1)^2 + \sin^2(2\pi\alpha k)}$$
$$ a_n = \prod_{k=1}^n\sqrt{2 - 2\cos^2(2\pi\alpha k)} $$
$$ a_n = 2^{\frac n2} \prod_{k=1}^n|\sin(2\pi\alpha k)|$$
Does this lead anywhere?