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Let $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ with continued fraction approximants $\frac{p_n}{q_n}$. So if $\alpha=[a_0;a_1,a_2;\dots]$, we have $q_n=a_n q_{n-1}+q_{n-2}$, with $q_0=1$ and $q_1=a_1$. Define $$ \beta(\alpha) = \limsup_{n\to \infty} \frac{\ln q_{n+1}}{q_n}.$$ It is said that $\alpha$ is an ``exponentially Liouville number'' if $\beta(\alpha)>0$.

This condition may have first appeared in this paper by Avila, but it may be older: https://arxiv.org/pdf/1006.0704.pdf

Now I had two questions. First, is the set of such numbers dense? I know that Liouville numbers are dense, but this condition seems stronger.

I believe they are dense. In fact, I believe that for any $M>0$, the set $\{\alpha : \beta(\alpha) = M\}$ is dense. For this, my heuristic is that given $x=[a_0;a_1,a_2,\dots]$, we can define $\alpha_n$ to agree with $x$ on a long string, then put somehing like $e^{Mq_{n-1}}$ on the $n$-the position (i.e. $a_n(\alpha_n) \approx e^{Mq_{n-1}}$). I am a greenhorn in this field, so can you confirm they are dense?

Next, can you tell me what is the relation between this condition and the ordinary definition of Liouville numbers? It should be clear that if $\beta(\alpha)>0$, then $\alpha$ is a Liouville number, but I can't find a good argument.

Thanks !

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    I always though that the set {$\frac{1}{n}$} , $n\in\mathbb N$ would be dense at $x=0$ , but I was surprised when I read that this is not the case because the interior of the set (I did not really understand what this is) is empty here. – 2017-02-13
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    Hello Peter. Thanks for your feedback. Your comment on irrationality measure here http://mathworld.wolfram.com/IrrationalityMeasure.html somehow answers my second question: if $\beta(\alpha)>0$, then $\mu(\alpha)=\infty$, so by the link, the number is Liouville. – 2017-02-13
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    Hello, I noticed that in the denominator there is no $\ln$. How are the two expressions related ? In particular, how can I use the new expression to prove that a number is transcendental (sufficient is to show that the irrationality measure is grater than $2$) ? – 2017-02-13
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    Yes, $\{\frac{1}{n}\}$ has empty interior and is not dense. There is no meaning for ``dense at $x$''. What you mean is that $x=0$ is a limit point of $\{\frac{1}{n}\}$, and this is true. – 2017-02-13
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    If $\beta(\alpha)>0$, then $\ln q_{n+1}$ is of order $q_n$ (at least on a subsequence), so $\frac{\ln q_{n+1}}{\ln q_n} \approx \frac{q_n}{\ln q_n} \to \infty$. – 2017-02-13
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    you can find the defition of the interior of a set on wikipedia. I suppose you can also find this in the baby Rudin, or any lecture notes on metric spaces. – 2017-02-13

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