TITLE: Is the $\sum\sin(n)/n$ convergent or divergent?
QUESTION [18 upvotes]: 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?

REPLY [24 votes]: 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, $$S_N = \frac{\cos \left( \frac1{2} \right) - \cos \left( N + \frac1{2} \right)}{2\sin \left(\frac1{2}\right)}$$