Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 20



Work Step by Step

In the given problem: $a_{n}=\frac{(-5)^{n}}{n^{2}9^{n}}=\frac{(25)^{n}}{n^{2}9^{n}}$ and $a_{n+1}=\frac{(25)^{n+1}}{(n+1)^{2}9^{n+1}}$ Re-write $a_{n+1}$ as $a_{n+1}=\frac{(25)^{n+1}}{(n+1)^{2}9^{n+1}}$ Therefore, $\frac{a_{n+1}}{a_{n}}= \frac{\frac{(25)^{n+1}}{(n+1)^{2}9^{n+1}}}{\frac{(25)^{n}}{(n)^{2}9^{n}}}$ $=\frac{25}{9}(\frac{n}{n+1})^{2}$ and $L=\lim\limits_{n \to \infty}|\frac{25}{9}(\frac{n}{n+1})^{2}|$ $=\frac{25}{9} \gt 1$ Thus, $L\gt 1$ Hence, the series is divergent by ratio test.
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.