Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 726: 27


The function is neither positive nor decreasing. Thus the integral test can not be applied.

Work Step by Step

Given: $\Sigma_{n=1}^{\infty}\frac{cos(\pi n)}{\sqrt n}$ The integral test requires a function that is positive and decreasing on $[1,\infty)$ . The $cos \pi n$ alternates between $-1$ and $+1$ , so it is not positive and decreasing for the entire interval. Thus the integral test can not be applied.
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