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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?

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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.

Is there some reason why you think the limit ought to be algebraic?

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  • $\begingroup$ 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 $x$? $\endgroup$ – user405683 Jan 16 '17 at 14:10
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Jan 16 '17 at 16:48
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jan 16 '17 at 21:57
  • $\begingroup$ @RobertIsrael thank you. How can one show that $\pi^2/(12 \log 2)$ is not the log of an algebraic number? $\endgroup$ – user405683 Jan 17 '17 at 8:43
  • $\begingroup$ @GerryMyerson thank you. That means, for algebraic $x$ of higher degree, we now nothing and there suspect that both cases could hold? $\endgroup$ – user405683 Jan 17 '17 at 8:44

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