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 $\pi$.
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5$\begingroup$ It's not a research level question. You should ask it at math.stack.exchange. $\endgroup$– Davide GiraudoCommented Jun 9, 2012 at 11:31
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In this particular case, use $s_n(\theta)=\sum_{k=0}^n\sin(k\theta)=\Im(\sum_{k=0}^n\exp(ki\theta))=\dots=\frac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin (\theta/2)}$
answered Jun 9, 2012 at 11:24
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1$\begingroup$ In particular, $s_n$ is bounded. $\endgroup$ Commented Jun 9, 2012 at 11:55