Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 745: 56


$a_{n}$ converges to $0$.

Work Step by Step

Use Ratio Test $\Sigma_{n=1}^{\infty}$ $a_{n}=\Sigma_{n=1}^{\infty} \frac{(-3)^{n}}{n!}$ $|a_{n}|=\frac{3^{n}}{n!}$ $\lim\limits_{n \to \infty} |\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty} |\frac{3^{n+1}}{(n+1)!}\times \frac{n!}{3^{n}}|= \lim\limits_{n \to \infty} |\frac{3n!}{(n+1)!}|$ By definition $(n+1)!=(n+1)n!$ $\lim\limits_{n \to \infty} |\frac{3n!}{(n+1)n!}| = \lim\limits_{n \to \infty} |\frac{3}{n+1}| = \frac{3}{\infty}= 0 \lt 1$ Thus, the series $\Sigma|a_{n}|$ is convergent by ratio test and by theorem $\lim\limits_{n \to \infty} |a_{n}|=0$ leads to $\lim\limits_{n \to \infty} a_{n}=0$ Therefore, $a_{n}$ converges to $0$.
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