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



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.
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