Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises: 3

Answer

Divergent

Work Step by Step

$\Sigma n^{-3}$ $\int_{1}^{\infty} (x^{-3})dx=\lim\limits_{t \to \infty}\int_{1}^{t} (x^{-3})dx=\lim\limits_{t \to \infty}[-\frac{1}{2}x^{-2}]_{1}^{t}=0--\frac{1}{2}=\frac{1}{2}$ $\frac{1}{2}<1$ Divergent
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