Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
up vote 4 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.