Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.1 An Overview - 8.1 Exercises: 46

Answer

Converges to $1$

Work Step by Step

$a_n = 2^nsin(2^{-n})$ $a_1 = 0.958851$ $a_2 = 0.989616$ $a_3 = 0.997398$ $a_4 = 0.999349$ $a_5 = 0.999837$ $a_6 = 0.999959$ $a_7 = 0.99999$ $a_8 = 0.999997$ $a_9 = 0.999999$ $a_{10} = 1.0$ The terms of the sequence seems to get closer and closer to $1$.
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