Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.6 Absolute Convergence and the Radio and Root Tests - 11.6 Exercises - Page 783: 14

Answer

Divergent

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.
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