Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 570: 65



Work Step by Step

As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ and $\lim\limits_{n \to \infty} \dfrac{x^n}{n!}=0$ Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n!}{10^{(6n)}}$ This implies that $ \lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{1}{\dfrac{((10)^{(6)})^n}{n!}}=\dfrac{1}{0}=\infty$ Thus, $\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is Divergent.
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