University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 46


Converges to zero.

Work Step by Step

Sine is in the interval $[-1,1]$. Its square is nonnegative and less (or equal to) 1. $0\leq\sin^{2}n\leq 1$ $\displaystyle \frac{1}{2^{n}}$ is positive and multiplying the compound inequality with it will not change the inequality direction. $0\displaystyle \leq\frac{\sin^{2}n}{2^{n}}\leq\frac{1}{2^{n}}$ Both $\{0\}$ and $\displaystyle \{\frac{1}{2^{n}}\}$ converge to zero. Applying the sandwich theorem, $\displaystyle \{\frac{\sin^{2}n}{2^{n}}\}$ also converges to zero.
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