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: 29



Work Step by Step

Since, the given series is a geometric series with common ratio $r=\dfrac{1}{\ln 2}$ Thus, for this series $|r|=\dfrac{1}{\ln 2} \approx 1.44 \gt 1$ Hence, the given series diverges.
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