Answer to Question #263433 in Calculus for Sia

Question #263433

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


1
Expert's answer
2021-11-15T20:22:59-0500

Recall that eiθ(cosθ+isinθ)Note thatn=1Neθnt=eiθ(1eiNθ)1eiθ21eiθ< since ei1 Since the sequence {1n}n=1 is a monotonically decreasing sequenceand limn1n=0. Hence, by the Dirichlet’s test we have that n=1einθn converges    s=n=1einθn=n=1cosnθ+isinnθn convergesConsequentially, we have thatIm(s)=n=1sinnθn convergesTaking θ=1, then we have our desired result.\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.}


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