Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises - Page 743: 14



Work Step by Step

Let $a_n=\frac{n!}{100^n}$. Then $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim\limits_{n \to \infty}\left|\frac{\frac{(n+1)!}{100^{n+1}}}{\frac{n!}{100^n}}\right|=\lim\limits_{n \to \infty}\left|\frac{(n+1)!}{100^{n+1}}\cdot\frac{100^n}{n!}\right|\lim\limits_{n \to \infty}\left|\frac{(n+1)!}{n!}\cdot\frac{100^n}{100^{n+1}}\right|=\lim\limits_{n \to \infty}\left|(n+1)\cdot\frac{1}{100}\right| =$ $$\frac{1}{100}\cdot\lim\limits_{n \to \infty}\left|n+1\right| =\infty.$$ Since $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty$, the series $\displaystyle{\sum_{n=1}^{\infty}\frac{n!}{100^n}}$ is divergent by the 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.