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 725: 61

Answer

The given sequence diverges.

Work Step by Step

$\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty} \frac{n^{2}cosn}{1+n^{2}}$ $=\lim\limits_{n \to \infty} \frac{\frac{n^{2}cosn}{n^{2}}}{\frac{1+n^{2}}{n^{2}}}$ $=\frac{\lim\limits_{n \to \infty}cosn}{\lim\limits_{n \to \infty} \frac{1}{n^{2}}+1}$ $=\frac{\lim\limits_{n \to \infty}cosn}{1}$ $\lim\limits_{n \to \infty}cosn$ does not exist because $cosn$ is a periodic function. Therefore, the given sequence diverges.
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