# limit of denominator in continued fraction expansion algebraic?

I asked this on MathStackExchange but didn't get an answer, so I'm trying it here:

Let $\alpha$ be an algebraic number and denote with $\frac{p_n}{q_n}$ the $n$-th convergent of $\alpha$ that we get when expressing $\alpha$ as continued fraction. Is it true that $\underset{n \rightarrow \infty}{\lim} \frac{1}{n} \log(q_n)$ is algebraic?

Let $\alpha$ be the golden ratio, $\phi$, an algebraic number. Then the $q_n$ are the Fibonacci numbers, thus asymptotic to $c\phi^n$ for some positive constant $c$, so the limit in question is $\log\phi$, which is not algebraic.
• Thanks for your answer. I was looking for counterexamples for Levys theorem (for almost all $x \in (0,1)$ we have $\underset{n \rightarrow \infty}{\lim} \frac{1}{n} \log(q_n) = \frac{\pi^2}{12 \log 2}$). Is this theorem always false for algebraic $x$? Jan 16 '17 at 14:10
• In the case of a quadratic irrational, we get an eventually periodic continued fraction, and the limit should be the log of an algebraic number, which $\pi^2/(12 \log 2)$ isn't. Jan 16 '17 at 16:48
• Very little is known about the continued fraction expansion of algebraic $x$ of degree 3 and up. In particular, I suspect it is consistent with current knowledge that they are all counterexamples to Levy, and equally consistent with current knowledge that they are all examples of Levy. Jan 16 '17 at 21:57
• @RobertIsrael thank you. How can one show that $\pi^2/(12 \log 2)$ is not the log of an algebraic number? Jan 17 '17 at 8:43
• @GerryMyerson thank you. That means, for algebraic $x$ of higher degree, we now nothing and there suspect that both cases could hold? Jan 17 '17 at 8:44