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I have been trying to solve the following problem : prove that, for any $R\in [0,1]$, there exists an $a\in \mathbb R\backslash \mathbb Q$ such that the radius of convergence of $\sum \frac{z^n}{\sin(an\pi)}$ is $R$.

It turns out that when a is algebraic, the radius is $1$. This can be seen using Thue-Siegel-Roth theorem for diophantine approximation, which allows us to show that the distances of the numbers $na$ to $\mathbb Z$ are sufficiently large for the series to converge.

I also managed to find an $a$ such that the radius is $0$, using a Liouville-style construction : let $a$ be the sum of reciprocals of the terms of the sequence $(u_n)$ defined by $u_1=2$ and $u_{n+1}=u_n^{u_n}$. $u_na$ can be written as an odd integer + $u_n\sum _{k=n+1}^\infty \frac{1}{u_k}$. As $\sum _{k=n+1}^{\infty} \frac{1}{u_k}\leqslant \frac{K}{u_{n+1}}$ where $K$ is a universal constant, we get $-\frac{x^{u_n}}{\sin(u_n\pi a)} \geqslant \frac{(xu_n)^{u_n}}{K\pi u_n}\to \infty$, and the radius is $0$.

However, I don't know how to proceed to get a radius $R\in ]0,1[$. This time we need to look at every term, we cannot focus only on a subsequence. Controlling $\limsup |\frac{1}{sin(an\pi)}|^{1/n}$ seems tricky due to the presence of the $1/n$.

I suppose there is a way to make an explicit construction to answer this question, but I guess there might be a more theoretical approach. Any help / hint will be greatly appreciated !

Tutut
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1 Answers1

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I believe I have a somewhat general strategy for finding constants with radii of convergence $R\in\left(0,1\right)$ using a similar approach to your Liouville-style construction, though there are still a few details that need to be worked out, mentioned at the end. Define two increasing integer sequences with initial values $u_1=g_1$ such that $u_{n+1}=u_n^{g_n}$. Then, $u_n=u_1^{\prod_{k=1}^{n-1} g_k}$, so every $u_k$ divides $u_n$ for $k\le n$. Then suppose we want the radius of convergence to be $R=\frac{1}{2}$, so we set $u_1=g_1=\frac{1}{R}=2$, where $R$ must be the reciprocal of an integer $\ge2$. Like in your construction, we get $u_na=c+\sum_{k=n+1}^\infty\frac{u_n}{u_k}$ where $a$ is the sum of the reciprocals of $u_n$ and $c$ is an integer, and $\sum_{k=n+1}^\infty\frac{u_n}{u_k}=\frac{u_n}{u_{n+1}}+\epsilon_{1,n}=u_n^{1-g_n}+\epsilon_{1,n}$ for some $0<\epsilon_{1,n}<<u_n^{1-g_n}$.

Now we get the $u_n$th term in the power series evaluated at $R$:

$\left|s_{u_n}\left(R\right)\right|=\left|\frac{R^{u_n}}{\sin\left(u_na\pi\right)}\right|=\frac{R^{u_n}}{\pi\left(u_n^{1-g_n}+\epsilon_{1,n}\right)-\epsilon_{2,n}}=\frac{1}{\pi}\frac{u_1^{-u_n}}{u_n^{1-g_n}+\epsilon_{3,n}}=\frac{1}{\pi}\frac{u_1^{-u_n}u_n^{g_n}}{u_n+u_n^{g_n}\epsilon_{3,n}}=\frac{1}{\pi u_n}\frac{\left(u_1^{-u_n/g_n}u_n\right)^{g_n}}{1+u_n^{g_n-1}\epsilon_{3,n}}=\frac{1}{\pi u_n\left(1+\epsilon_{4,n}\right)}\left(u_1^{\left(\prod_{k=1}^{n-1}g_k\right)-u_n/g_n}\right)^{g_n}$

for some $0<\epsilon_{2,n}<<\pi\left(u_n^{1-g_n}+\epsilon_{1,n}\right)\approx u_n^{1-g_n}$ and $\epsilon_{3,n}=\epsilon_{1,n}-\frac{\epsilon_{2,n}}{\pi}$ where $|\epsilon_{3,n}|<<u_n^{1-g_n}$ and $\epsilon_{4,n}=u_n^{g_n-1}\epsilon_{3,n}$ where $|\epsilon_{4,n}|<<1$.

Now we can solve for $\left(\prod_{k=1}^{n-1}g_k\right)-u_n/g_n=0$, so $\left(u_1^{\left(\prod_{k=1}^{n-1}g_k\right)-u_n/g_n}\right)^{g_n}=1$, which will cause $s_{u_n}\left(R\right)$ to approach $0$ due to the $\frac{1}{\pi u_n\left(1+\epsilon_{4,n}\right)}$ term, thus converging at $R$. Rearranging, we get $u_n=\prod_{k=1}^ng_k=u_{n-1}^{g_{n-1}}$ using the definition of $u_n$, and so $g_n\prod_{k=1}^{n-1}g_k=u_{n-1}^{g_{n-1}}$ which becomes $g_{n+1}=\frac{u_n^{g_n}}{\prod_{k=1}^ng_k}$. Now we can calculate all values of $u_n$ and $g_n$. To test that this $R$ is an upper bound, we can use $\frac{1}{u_1-\epsilon_5}$ for some $0<\epsilon_5<u_1-1$ for the value of the power series instead:

$\left|s_{u_n}\left(\frac{1}{u_1-\epsilon_5}\right)\right|=\frac{1}{\pi u_n\left(1+\epsilon_{4,n}\right)}\left(\left(u_1-\epsilon_5\right)^{-u_n/g_n}u_n\right)^{g_n}=\frac{1}{\pi u_n\left(1+\epsilon_{4,n}\right)}\left(u_1^{\left(\prod_{k=1}^{n-1}g_k\right)-\log_{u_1}\left(u_1-\epsilon_5\right)u_n/g_n}\right)^{g_n}=\frac{1}{\pi u_n\left(1+\epsilon_{4,n}\right)}\left(u_1^{u_n/g_n\left(1-\log_{u_1}\left(u_1-\epsilon_5\right)\right)}\right)^{g_n}=\frac{1}{\pi u_n\left(1+\epsilon_{4,n}\right)}\left(\frac{u_1}{u_1-\epsilon_5}\right)^{u_n}\approx\frac{\left(1+\epsilon_5/u_1\right)^{u_n}}{u_n}\to\infty$

The first issue with this approach is that it doesn't account for the possibility that the constant $a$ has significantly better rational approximations than the obvious ones given by the series, I am not sure if this has an easy solution.

The second issue is, if $R$ isn't the reciprocal of an integer $\ge2$ we run into issues with the sequences not being integer sequences, thus not forcing the values of $\sin(u_na\pi)$ to be near $0$. However, it is possible that the approach could be modified, maybe using floor/ceil/round functions in the and/or a third sequence, but I have yet to figure out anything worth pursuing.

As a side note, for any two sequences $u_{1,n}<u_{2,n}$ with different initial values, there is some fixed $k$ for which $u_{2,n}<u_{1,n+k}$ is always true, so in some sense the growth rates of these sequences are all the same. Also, any $a$ satisfying the conditions has to be Liouville, as otherwise the $R^n$ term will always eventually dominate any denominator $n^\mu$ for irrationality exponent $\mu$; in fact, $a$ has irrationality base $u_1=\frac{1}{R}$. A good followup question which seems doable would be whether irrationality base and radius of convergence are interchangeable, which could help solve the second issue. I am already convinced that they are indeed interchangeable, but this answer is long enough as it is.

Let me know if I missed anything!

David Chew
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