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.
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Sign up to join this communityDoes $\sum\frac{\sin n}{n}$ converge?
I have tried the comparison test, root test and ratio test but still can't prove it is convergent or divergent.
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 $$ |S_n| \leq \left|e^i\frac{1-e^{in}}{1-e^i}\right| \leq \frac{2}{|1-e^i|} < \infty. $$
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 $$\sum_{k=1}^{n}\sin kx=\frac{\cos\left(\frac{1}2x\right)-\cos\left(n+\frac{1}2\right)x}{2\sin(x/2)},x\neq0,\pm\pi,\pm 2\pi,...$$
$\displaystyle \sum \frac{\sin n}{n}$ converges