Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 538: 57

Answer

The sequence $a_n$ converges to $\pi$.

Work Step by Step

We have $$\lim_{n\to \infty}a_n=\lim_{n\to \infty}n \sin \frac{\pi}{n} \\ = \pi\lim_{n\to \infty}\frac{ \sin \frac{\pi}{n} }{\pi/n}= \pi\lim_{\pi/n\to 0}\frac{ \sin \frac{\pi}{n} }{\pi/n}= \pi.$$ Here, we used the fact that $\lim_{x\to 0}\frac{ \sin x}{x}=1$. Hence, by Theorem 1, the sequence $a_n$ converges to $\pi$.
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