Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 636: 5



Work Step by Step

Consider $a_n=\dfrac {\sin n\pi}{2}$ The given sequence diverges because $\dfrac {\sin n\pi}{2}=0,1,0,-1,0,1,0,-1,0,1,0-1.....$so on This means the the given sequence does not have a finite value. Hence, the sequence is divergent.
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