Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 748: 11

Answer

The series is convergent. $\sum_{n=1}^\infty\frac{6}{(-3)^n}=-1.5$

Work Step by Step

The graph of the sequence of terms and the sequence of partial sums is given below. From the graph, the sequence of partial sums $S_n$ approaches $-1.5$ as $n$ becomes larger, so that the series is convergent. The series $\sum_{n=1}^\infty\frac{6}{(-3)^n}$ is an infinite geometric series with the first term $a_1=-2$ and the common ratio $r=-\frac{1}{3}$. Find the sum of the series: $\sum_{n=1}^\infty\frac{6}{(-3)^n}=\frac{a_1}{1-r}=\frac{-2}{1-(-\frac{1}{3})}=\frac{-2}{4/3}=-1.5$
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