Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.1 Exercises - Page 724: 36

Answer

Convergent with limit 1

Work Step by Step

We are given the sequence: $a_n=\cos\left(\dfrac{2}{n}\right)$. As $n$ increases, $\dfrac{2}{n}$ decreases, so $\cos\left(\dfrac{2}{n}\right)$ increases; therefore $a_n$ is a monotonic sequence. On the other hand, $\cos\left(\dfrac{2}{n}\right)$ is a bounded sequence because $\cos x$ is a bounded sequence. So, according to the Monotonic Sequence Theorem, the sequence $a_n$ is convergent. Calculate the limit: $\displaystyle{\lim_{n \to \infty}} \cos\left(\dfrac{2}{n}\right)=\cos \left(\displaystyle{\lim_{n \to \infty}}\left(\dfrac{2}{n}\right)\right)=\cos 0=1$ Therefore the sequence converges and its limit is 1.
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