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 497: 9

Answer

Diverges

Work Step by Step

Write the series as a sum of two series $\Sigma_{n=1}^{\infty}(1-\dfrac{7}{4^n})=\Sigma_{n=1}^{\infty}1-\Sigma_{n=1}^{\infty}7(\dfrac{1}{4^n})$ Here, $-\Sigma_{n=1}^{\infty}7(\dfrac{1}{4^n})$ converges because it a geometric series with common ratio $r=\dfrac{1}{4}\lt 1$ But $\Sigma_{n=1}^{\infty}1$ diverges by the n-th term test. Thus, the final series will be divergent.
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