University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 29

Answer

Diverges

Work Step by Step

Consider $a_n=\dfrac{1}{\sqrt n \ln n}$ When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase. Thus, we have $a_n \geq \dfrac{1}{\sqrt n \sqrt n}=\dfrac{1}{n}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent series. Thus the series diverges by the direct comparison test.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.