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 - 11.4 Exercises - Page 751: 45

Answer

$\Sigma sin(a_{n})$ converges.

Work Step by Step

Yes it is true. Since $\Sigma a_{n}$ converges, $\lim\limits_{n \to \infty}a_{n}=0$ But $sinx\lt x$, for all $x\gt 0$ , so $a_{n}\gt sin (a_{n})$ for all $n$ . Thus, by the comparison test $\Sigma a_{n}$ converges, so $\Sigma sin(a_{n})$ converges.
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