Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 24

Answer

This geometric series diverges.

Work Step by Step

Since the given series is in a similar form as the definition of the geometric series, we can make it very close by taking one 3 outside, i.e. $3^{n+1}$ becomes $3^{n}$. The summation of an infinite geometric series is $S= \frac{a}{1-r}$ if |$r$| $\lt 1 $ (here $r$ is the common ratio of the geometric series). Now, If we choose n = 0 we can see that the initial term is $a$ = $3$ and the common ratio is $r$ = $\frac{-3}{2}$ (just by looking at the form in the definition and the one in the question after factoring out a $3$). $r$ = |$\frac{-3}{2}$| $\gt 1 $ , which means the series is divergent.
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