6
$\begingroup$

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?

  • 0
    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.2011-07-02

1 Answers 1

4

$\sin(\log(k^m)) = \sin(m\;\log(k))$ 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.