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Possible Duplicate:
Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$

So, in my calculus class (one I'm teaching, not taking), the sum $\sum_{n=1}^\infty \frac{\sin(n)}{n}$ has come up a few times. Unfortunately, as someone not exactly an expert in the convergence of sums, it seems to resist the few techniques I know. Certain none of the usual first year calculus tricks (integral test, alternating series test, ratio test, etc.) work, and the only more tricky technique, partial summation, I can think of doesn't seem to work either (one would need that $\sum_{n=1}^N\sin(n)$ is bounded, which I believe is false).

It seems like it should converge, since it switches sign quite often, but on the other hand, the harmonic series can mess with your intuition, so I don't have much trust in that. So, I ask to you:

Does this series converge?

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    @RossMillikan: Not necessarily, the statement may be undecidable ( not likely in this case, but, in theory, possible).2017-12-10

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The sum of $\sum_{n=1}^{N} \sin(n) = \frac{\sin(N) - \cot \left( \frac1{2} \right) \cos \left( N \right) + \cot \left( \frac1{2} \right)}{2}$ which is clearly bounded and hence by generalized alternating series test (also known as Dirichlet's test) the sum converges.

EDIT $S_N = \sum_{n=1}^{N} \sin(n)$

$2\sin \left( \frac1{2} \right) \times S_N = \sum_{n=1}^{N} \left( \cos \left( n- \frac1{2}\right) - \cos \left( n+ \frac1{2}\right)\right) = \cos \left( \frac1{2} \right) - \cos \left( N + \frac1{2} \right)$

Hence, $S_N = \frac{\cos \left( \frac1{2} \right) - \cos \left( N + \frac1{2} \right)}{2\sin \left(\frac1{2}\right)}$

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    @ShreevatsaR: $S_N$ doesn't need to have a limit. All we need is $S_N$ to be bounded. (In fact consider the alternating series $1- \frac1{2} + \frac1{3} - \frac1{4} \ldots$. The sequence of partial sums of the numerator is $1,0,1,0,\ldots$ which doesn't have a limit but it is bounded and hence the series converges.)2011-05-03