Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 31


The given series is convergent and its is 9.

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- (by only factoring out a $3$ from the series and thus we will get: $3 * \Sigma (\frac{3}{4})^{n} $ 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 $a$ = $\frac{9}{4}$ and the common ratio $r$ = $\frac{3}{4}$ (Just by dividing any latter term by its former). |$r$| = $\frac{3}{4}\lt 1 $, which means that the series is convergent. Using: $S= \frac{a}{1-r}$ = $\frac{\frac{9}{4}}{1-\frac{3}{4}}$ = $\frac{\frac{9}{4}}{\frac{1}{4}}$ = 9. The given series is convergent and its is 9.
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