MathOverflow is a question and answer site for professional mathematicians. 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 wonder whether the sequence $s_{n} = \sum_{k = 1}^{n} \sin k$ is bounded.

The answer seems no, but I have no idea how to prove this from the irrationality of $\pi$.

share|cite|improve this question

closed as off topic by Gerry Myerson, George Lowther, Felipe Voloch, Mark Meckes, Vladimir Dotsenko Jun 9 '12 at 13:28

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

It's not a research level question. You should ask it at – Davide Giraudo Jun 9 '12 at 11:31

In this particular case, use $s_n(\theta)=\sum_{k=0}^n\sin(k\theta)=\Im(\sum_{k=0}^n\exp(ki\theta))=\dots=\frac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin (\theta/2)}$

share|cite|improve this answer
In particular, $s_n$ is bounded. – David Speyer Jun 9 '12 at 11:55
Thanks for your help! – Vincent Jun 9 '12 at 12:23

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