Chapter 8.3, Problem 29E

To determine

The series is convergent or divergent.

Answer to Problem 29E

The series is convergent.

Explanation of Solution

Given information: The series is ${\displaystyle \sum _{n=1}^{\infty }\sin \frac{1}{n}}$

Theorem: The necessary condition for convergence of the series ${\displaystyle \sum u_n}$ is that $\underset{x\to \infty }{\lim }{u}_n\to 0\text{ as }n\to \infty .$

Given, ${\displaystyle \sum _{n=1}^{\infty }\sin \frac{1}{n}}$

Let, $u_k=\cos \frac{1}{n}$

  $\Rightarrow u_k=\sin \frac{1}{n}\le \frac{1}{n}=v_n(say)$

  ${\displaystyle \sum _{n=1}^{}{v}_n}$is a geometric series where $\frac{1}{n}\ge 1$ .so the series ${\displaystyle \sum _{n=1}^{}{v}_n}$is convergent

Since, the series ${\displaystyle \sum v_n}$ is divergent and so by the comparision test. The series ${\displaystyle \sum u_n}$ is convergent .

Hence , the series is convergent.