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: 17


If we look at the ratio change from term to term we find that: 3$* \frac{-4}{3}$= -4 , and -4 * $\frac{-4}{3}$ = $\frac{16}{3}$ and so on, thus we have $r = \frac{-4}{3}$ looking at the formulas and definition 4. in page 750, we find that $ | \frac{-4}{3} | \geq1$. Result: the series diverges.

Work Step by Step

Since we are given the information in the question that the series is geometric, we look for the common ratio. To do that, we can divide the terms. For example: $\frac{-64}{9} \div \frac{16}{3} = \frac{-4}{3} $ Same for the third term divided by the second: $\frac{16}{3} \div -4 = \frac{-4}{3}$. Then, after finding the ratio, we look at definition 4 and its formulas in page 750 and we observe that $ | r | = \frac{4}{3} \geq1$. which means that the series diverges (which, for explanation, means that the series sum goes to infinity as the number of terms increase)
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.