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I tried (and still try) to prove that $\sin (\log n)$ doesn't have a limit at $\infty$. I know it is enough to show that a subsequence of $\log n$ approaches, modulo $2\pi$, arbitrarily close to 2 distinct values $\alpha, \beta$ (such that $\sin \alpha \neq \sin \beta$), which is a much weaker statement than "$\ \frac{\log n}{2\pi}$ is equidistributed modulo $1$" (which is even false...).

My questions are:

  1. How do you prove my specific case?

  2. Is there a general theory of limit of $f \circ g (n)$ where $f$ is a periodic continuous function and $g$ is continuous, (possibly monotone increasing) function diverging to $\infty$ at $\infty$?

  3. What is a good source about equidistribution?

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5  
All you need to know is that $\log n$ goes to $\infty$ but that $|\log(n+1) - \log n|$ tends to zero. –  Qiaochu Yuan Jul 1 '11 at 23:00
2  
I think the go-to source on equidistribution is Kuipers and Niederreiter, "Uniform Distribution of Sequences". –  John M Jul 2 '11 at 4:25
    
Try computing and simplifying $\sin (\log (n+1)) - \sin(\log(n))$ as $2 \cos A \sin B$ - the $cos$ factor is bounded and you should be able to prove that the $sin$ factor is small. –  Mark Bennet Jul 2 '11 at 10:12

1 Answer 1

up vote 3 down vote accepted

$\sin(\log(k^m)) = \sin(m\;\log(k))$ is a subsequence of the form $\sin(m\alpha)$ and may be easier to work with. The values of $\alpha$ where the sequence $\sin(m\alpha)$ is convergent have to be pretty rare.

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