Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.3 - The Integral Test - Exercises 10.3 - Page 586: 46

Answer

NO

Work Step by Step

As we can see that $(\dfrac{1}{2})\Sigma_{n=1}^\infty a_n$ is a convergent series of positive numbers. Thus, $(2)\Sigma_{n=1}^\infty a_n= \Sigma_{n=1}^\infty 2a_n$ also diverges for $2a_n \geq a_n$ Therefore, we will not get the largest convergent series of positive numbers. Thus, the correct answer is No.
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.