Answer to Question #263433 in Calculus for Sia

Question

Convergence test for $\displaystyle\sum_{n=1}^\infty \frac{sin(n)}{n}$.

Answer 1

$\displaystyle \text{Recall that $e^{i\theta}(\cos \theta$}+i\sin\theta)\\ \text{Note that}\\ \left|\sum^N_{n=1}e^{\theta n t}\right|= \left|\frac{e^{i\theta}(1-e^{iN\theta})}{1-e^{i\theta}}\right|\leq \frac{2}{|1-e^{i\theta}|}<\infty\text{ since $e^i \neq 1$ }\\ \text{Since the sequence $\{\frac{1}{n}\}^\infty_{n=1}$ is a monotonically decreasing sequence}\\ \text{and $\displaystyle\lim_{n\to \infty}\frac{1}{n}=0$. Hence, by the Dirichlet's test we have that }\\ \sum^\infty _{n=1}\frac{e^{in\theta}}{n} \text{ converges}\\ \implies s = \sum^\infty _{n=1}\frac{e^{in\theta}}{n} = \sum^\infty _{n=1}\frac{\cos{n\theta}+i\sin{n\theta}}{n} \text{ converges}\\ \text{Consequentially, we have that}\\ Im(s) = \sum^\infty_{n=1}\frac{\sin{n\theta}}{n} \text{ converges}\\ \text{Taking $\theta = 1$, then we have our desired result.}$