Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.4 - The Comparison Tests - 11.4 Exercises - Page 731: 30

Answer

Convergent

Work Step by Step

To use the Direct Comparison Test, we need a known series $\Sigma_{n=1}^{\infty}b_{n}$ for the purpose of comparison. We will choose $b_{n}=\frac{2}{n^{2}}$ The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Now, $a_{n}=\frac{n!}{n^{n}}$ and $b_{n}=\frac{2}{n^{2}}$; which is p-series with $p =2\gt 1$ Here, $a_{n} \lt b_{n}$ for $n\geq 4$ , because $\frac{1}{n}\times \frac{2}{n} \times [\frac{3}{n} \times \frac{4}{n}...\frac{n}{n}] \lt \frac{1}{n} \times \frac{2}{n}$ Therefore, $\Sigma a_{n}$ is convergent by comparison test. Hence, the $\Sigma_{n=1}^{\infty}\frac{n!}{n^{n}}$ series is convergent.
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