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: 23

Answer

The answer is: This geometric series converges and its sum is 17 .

Work Step by Step

Since the given series is in the same form as the definition of the geometric series that can be found in page 750 definition 4 -second line in the page- (except for the power of $r$ being n instead of $n-1$). 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 $a$ = $\frac{1}{4}$ and the common ratio $r$ = $\frac{-3}{4}$ (Dividing any latter term by its former we can get the common ratio). $r$ = |$\frac{-3}{4}$| $\lt 1 $ , which means the series is convergent. Now we only substitute in $S= \frac{a}{1-r}$. $S= \frac{\frac{1}{4}}{1-\frac{-3}{4}}$ = $S= \frac{\frac{1}{4}}{\frac{7}{4}}$ = $\frac{1}{7}$ . The answer is: This geometric series converges and its sum is $\frac{1}{7}$ .
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.