## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 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 that can be found in page 750 definition 4 -second line in the page- (we can make it very close by taking one 3 outside so $3^{n+1}$ 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 $a$ = $3$ and the common ratio $r$ = $\frac{-3}{2}$ (Just by looking at the form in definition and the one in the question after factoring out a $3$). $r$ = |$\frac{-3}{2}$| $\gt 1$ , which means the series is divergent. The answer is: This geometric series diverges.

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