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: 48


$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.

Work Step by Step

Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}$ But $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\dfrac{\infty}{\infty}$ The limit shows an Indeterminate form. So, we will use L-Hospital's rule. , $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\lim\limits_{n \to \infty} \dfrac{3^n \ln 3}{3n^2}$ or, $\lim\limits_{n \to \infty} \dfrac{3^n \ln 3}{3n^2}=\dfrac{\infty}{\infty}$ Apply L-Hospital's rule again. $\lim\limits_{n \to \infty} \dfrac{3^n (\ln 3)^3}{(6)}=\infty$ Thus, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
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