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I am studying functions given by the power series: $$f(z)=1+\sum_{n=1}^{\infty}\frac{z^n}{(1-q)(1-q^2)\cdots(1-q^{n})}.$$ The parameter $q$ is usually assumed to be such that $|q|<1$. Then it is easy to see that the radius of convergence for the above series is equal to 1.

However, my interest concerns the boundary case when $|q|=1$ and $q$ is not a root of unity (to not to divide by $0$). Then one can show that the radius of convergence $R=R(q)$ (a non-trivial statement) is given by $$R(q)=\liminf_{n\to\infty}\left|1-q^{n}\right|^{1/n}.$$

If we write $q=e^{2\pi i \omega}$ then the question can be reformulated in terms of diophantine properties of the number $\omega$. Since $|1-q^{n}|=2|\sin(\pi(n\omega-k))|$ for an arbitrary integer $k$, we get $$R(q)=\liminf_{n\to\infty}\|n\omega\|^{1/n}$$ where $\|x\|$ denotes the distance from $x$ to the nearest integer.

Here the number theory comes into the game. I found in a book without reference or proof that $R(q)=1$ for almost every $q$ on the unit circle.

Q1: Can you provide a reference for the above statement?

Further, I would like to understand a little bit more the structure of the set $$\mathcal{R}=\{\omega \mid R(q)=1\}.$$ For example, with a given irrational number $\omega$ (e.g. $\pi$, $e$, $\tau$, etc.) is there a way how to decide whether $\omega\in\mathcal{R}$?

Q2: What irrational numbers belong to $\mathcal{R}$?

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    $\begingroup$ Almost all real numbers $\omega$ has irrationality measure 2, that is, $\|n\omega\|\cdot n^c\rightarrow \infty$ for any fixed $c>1$ and large $n$. This is much stronger that $\|n\omega\|^{1/n}\to 1$. $\endgroup$ Jan 29, 2016 at 13:57
  • $\begingroup$ About the "non-trivial statement", is it a consequence of Cauchy-Hadamard's theorem? $\endgroup$ Mar 20, 2020 at 21:48

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