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

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

1 Answer 1

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

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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

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