Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 758: 85


$\Sigma(a_{n} +b_{n})$ will be divergent.

Work Step by Step

Let us assume $\Sigma (a_{n} + b_{n})$ is convergent . It is given that $\Sigma a_{n}$ is convergent. Theorem 8 states that : $\Sigma^{\infty}_{n=1} (a_{n} + b_{n}) = \Sigma^{\infty}_{n=1}a_{n} + \Sigma^{\infty}_{n=1}b_{n}$ By Theorem 8, we can say that $\Sigma(a_{n} + b_{n})-a_{n}=\Sigma b_{n}$ converges. Which contradicts the given information, which says that $\Sigma b_{n}$ diverges. Hence, $\Sigma(a_{n} +b_{n})$ will be divergent.
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