Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 8



Work Step by Step

Given:$a_{n}=\frac{(-10)^{n}}{n!}$ A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant. Remember that: $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ Therefore, $\sum_{n=0}^{\infty}\frac{(-10)^{n}}{n!}=e^{-10}$ Apply Divergence Test for Series. The series $\Sigma a_{n}$ converges if and only if $\lim\limits_{n \to \infty} a_{n}=0$ Since the series $\Sigma a_{n}$ converges where $a_{n}=\frac{(-10)^{n}}{n!}$ We conclude by the divergence test that $\lim\limits_{n \to \infty} a_{n}=\lim\limits_{n \to \infty} \frac{(-10)^{n}}{n!}=0$ Because the limits exists , the sequence converges. Hence, the given sequence converges to $0$.
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