University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 19

Answer

Converges

Work Step by Step

Consider $a_n=\dfrac{\sin ^2 n}{2^n}$ We know that $0 \leq \sin^2 n \leq 1$ Now, $a_n \leq \dfrac{1}{2^n}$ and $\Sigma_{n=1}^\infty \dfrac{1}{2^n}$ is a geometric convergent series with $r=\dfrac{1}{2}$ Hence, the series converges due to the direct comparison test.
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