Calculus: Early Transcendentals (2nd Edition)

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

Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 638: 33



Work Step by Step

$\frac{1}{\sqrt[3] k}=\frac{1}{k^{\frac{1}{3}}}$ $p=\frac{1}{3}\lt1$, therefore the series diverges.
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