University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 85


See below.

Work Step by Step

Consider $a_n=(\dfrac{1}{4})^n; b_n=(\dfrac{1}{2})^n$ Let $A=\Sigma_{n=1}^{\infty}a_n=\dfrac{1}{3}$ and $B=\Sigma_{n=1}^{\infty}b_n=1$ and $\Sigma_{n=1}^{\infty}(\dfrac{a_n}{b_n})=\Sigma_{n=1}^{\infty}(\dfrac{1}{2})^n=1 \ne \frac{A}{B}$
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