Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

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?

share|cite|improve this question

marked as duplicate by t.b., Zev Chonoles, Aryabhata, Ben Webster, Isaac May 4 '11 at 23:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Andrey Rekalo's answer here seems to answer your question. – t.b. May 3 '11 at 17:38
According to Wolphram Alpha it converges to $\frac{1}{2}(\pi-1)$ – Américo Tavares May 3 '11 at 17:42
@Américo: in the thread I linked to there is a derivation of that limit. – t.b. May 3 '11 at 17:44
@Theo: I started writing my comment before yours was visible. Now it is irrelevant. – Américo Tavares May 3 '11 at 17:50
Given the way the question is titled, I would answer "yes" – Ross Millikan Oct 10 '11 at 13:00

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

share|cite|improve this answer
What does the Generalized Alternating Series Test say? – Jonathan Gleason May 3 '11 at 17:42
@GleasSpty: The generalized alternating test is also known as Dirichlet's test. Wiki explains this in some detail – user17762 May 3 '11 at 17:46
How does one prove that formula for the sum? I'm guessing that writing $sin(n) = (e^{in} - e^{-in})/(2i)$ and then following your nose gives the result, but I haven't tried it. – Michael Lugo May 3 '11 at 17:53
@Michael: An elegant way in my opinion would be to multiply the sum by $\sin( \frac1{2})$ and then write it as a telescopic difference of cosines canceling out the terms to get the answer in the final form. – user17762 May 3 '11 at 17:54
@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.) – user17762 May 3 '11 at 18:16

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