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: 55



Work Step by Step

Given: $a_n=\frac{n!}{2^{n}}$ $\lim\limits_{n \to \infty}\frac{n!}{2^{n}}=\lim\limits_{n \to \infty}\frac{1\times 2\times 3\times 4\times ....\times (n-1)\times n}{2\times 2 \times 2 \times 2\times 2\times ....\times2 \times 2}$ $=\lim\limits_{n \to \infty}\frac{1}{2}\times \frac{2}{2}\times \frac{3}{2} \times ....\times \frac{n-1}{2} \times \frac{n}{2}$ $=\lim\limits_{n \to \infty}\frac{1}{2}\times \frac{2}{2}\times [\frac{3}{2} \times ....\times \frac{n-1}{2} \times \frac{n}{2}]$ The bracket term part is a product of infinitely many terms , each of which is greater than $1.5$ after the first term. Therefore, the bracket term part is greater than $1.5^{\infty}=\infty$ As we know $a^{\infty}=\infty$ if $a\gt 1$ Thus, $=\lim\limits_{n \to \infty}\frac{1}{2}\times\frac{2}{2}\times [\infty ]$ $=\infty $ The sequence diverges.
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