University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 21

Answer

Converges

Work Step by Step

The given series is a geometric series with common ratio $r=\dfrac{2}{3}$ Thus, for this series $|r|=\dfrac{2}{3} \lt 1$ Hence, the given series converges.
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