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?
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.