Does $\sum\frac{\sin n}{n}$ converge? Does $\sum\frac{\sin n}{n}$ converge?
I have tried the comparison test, root test and ratio test but still can't prove it is convergent or divergent.
 A: $\displaystyle \sum \frac{\sin n}{n}$ converges


*

*$\displaystyle \sum_{n = 1}^M \sin n$ is bounded always (why?)

*$\dfrac{1}{n} \to 0$ and is decreasing as $n \to \infty$

*We can conclude by Dirichlet's Test for convergence, which is sort of a generalization of the 'alternating series test,' that our series converges.

*This convergence is not absolute. In particular, $\max\{|\sin n|, |\sin (n + 1)|\}> \delta > 0$ for some $\delta$ (not depending on $n$), and so combining every two adjacent terms of the absolute series will diverge by comparison with the harmonic series.

A: Hint: I found this hint from my old notes. I hope it helps you.

$$\sum_{n=1}^{\infty}\frac{\sin nx}{n}$$ is uniformly convergent in any interval which doesn't include $$0,\pm\pi,\pm 2\pi,...$$ 

To see this use the Dirichlet test and this fact that $$\sum_{k=1}^{n}\sin kx=\frac{\cos\left(\frac{1}2x\right)-\cos\left(n+\frac{1}2\right)x}{2\sin(x/2)},x\neq0,\pm\pi,\pm 2\pi,...$$
A: You can use Dirichlet's test: the sequence $\frac{1}{n}$ is decreasingly converging to $0$, so you have to prove that
$$
S_n=\sum_{k=1}^n \sin k
$$
is bounded.
Here is a quick way to prove it: using $S_n = \Im(\sum_{k=1}^n e^{ik})$ and the inequality $|\Im(z)| \leq |z|$, we have
$$
|S_n| \leq \left|e^i\frac{1-e^{in}}{1-e^i}\right| \leq \frac{2}{|1-e^i|} < \infty.
$$
