Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 759: 32

Answer

Convergent for $p\gt 1$.

Work Step by Step

$\int_{3}^{\infty}\frac{1}{xlnx[ln(lnx)]^{p}}dx=\int_{ln(ln3)}^{\infty}\frac{1}{u^{p}}du=\int_{ln(ln3)}^{\infty}u^{-p}du$ which is convergent for $p\gt 1$.
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