University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 45

Answer

Yes

Work Step by Step

The answer is YES. Because, we have $\frac{1}{2}\Sigma_{n=1}^\infty a_n= \Sigma_{n=1}^\infty\dfrac{a_n}{2}$ also diverges and $\dfrac{a_n}{2} \lt a_n$ If we take a divergent series and divide each term by 2, we will still get a divergent series with smaller terms. Hence, the answer is YES.
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