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: 73


converges to $0$

Work Step by Step

As we know that when $x \gt 0$, $\lim\limits_{n \to \infty} \dfrac{x^n}{n!}=0$ Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}$ This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}$ and $=\lim\limits_{n \to \infty} \dfrac{(3^n 6^n 2^n)}{n!}=\dfrac{36^n}{n!}=0$ Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} converges to $0$.
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