# Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 45

Diverges

#### Work Step by Step

Let $u_n= \sin (\dfrac{1}{n})$ and $v_n=(\dfrac{1}{n})$ Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{ \sin (\dfrac{1}{n})}{\dfrac{1}{ n}}=\dfrac{\infty}{\infty}$ Need to apply L-Hospital's rule. $\implies \lim\limits_{n \to \infty} \dfrac{\cos (1/n)}{1}= \cos (0)=1 \ne 0 \ne \infty$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(1)}}$ is a divergent $p$-series test with $p \leq 1$ Hence, the series diverges by the limit comparison test.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.