## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 25

#### Answer

This geometric series diverges.

#### Work Step by Step

The given series can be made into a similar form as the definition of the geometric series found in page 750 definition 4 -second line in the page- (we can reach the textbook form by inputting $n=1$ to get the initial term $a$ and see the factor difference when inputting $n=2$). Note that 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 = 1 we can see that the initial term $a$ = $e^{2}$ and the common ratio $r$ = $\frac{e^{2}}{6}$ (Just by dividing any latter term by its former). |$r$| = $\frac{e^{2}}{6}\approx 1.232 \gt 1$ Thus, this geometric series diverges.

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