An irrational number $u\in(0,1)$ is said to be very-well approximable if there exists some $\delta>0$ for which there are infinite pairs of integers $p,q$ with $(p,q)=1$ (they are coprime) and

$$\left| u-\frac{p}{q}\right|<\frac{1}{q^{2+\delta}}$$

It is relatively easy to see that this is equivalent to the existence of some $\delta>0$ for which $a_{n+1}\ge q_n^\delta$ for infinitely many values of $n$, where $a_{n+1}$ is the $(n+1)$th term in the continued fraction of $u$, denoted by $u=[0,a_1,a_2,\dots]$, with $a_i\in\mathbb{Z}^+$.

I'd like to show that, for any well approximable number, the following convergence fails to hold

$$ \lim_{n\to\infty} \frac{1}{n} \log q_n=\frac{\pi^2}{12\log 2}$$

where $p_n/q_n$ are the convergents associated to $u$, that is, the finite continued fraction $\frac{p_n}{q_n}=[0,a_1,\dots,a_n]\in\mathbb{Q}$.

I'm thinking this must be rather elementary, but I can't seem to give a proof. Any hints?

In is known that the limit equality above holds for almost every number in $(0,1)$ (taking the corresponding convergent), but the proof of that is quite involved, and going through it I don't see why it should fail when $a_{n+1}\ge q_n^\delta$ infinitely often. Any hints?

  • $\begingroup$ How do you show that $u$ is very well approximable implies $a_{n+1}\ge q_n^\varepsilon$ for infinitely many values of $n$? I can only figure this out when the denominators of infinitely many rational numbers $q$ are actually $q_n$. But how do you show the general case? $\endgroup$
    – Bach
    Commented Feb 7, 2020 at 21:36
  • $\begingroup$ Okay, I have figured it out. We can express $q\in (q_{n-1},q_n]$ as $aq_{n-1}+bq_n$, for $a,b\in\mathbb Z$ and invoke the fact that $\left| u-\frac{p_n}{q_n} \right|>\frac{1}{(a_{n+1}+2)q_n^2}$. $\endgroup$
    – Bach
    Commented Feb 7, 2020 at 22:02

1 Answer 1


If $\frac1n \log q_n$ converges to any positive constant, then it is elementary that $\log q_{n+1}/\log q_n \to 1$. So then we cannot have $q_{n+1} > q_n^{1+\delta}$ infinitely often.

  • 1
    $\begingroup$ Well, this is embarrassing, I can't believe I didn't think of that. Thank you very much! $\endgroup$
    – Reveillark
    Commented Jun 15, 2017 at 11:47

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