University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 48


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

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}$ But $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\dfrac{\infty}{\infty}$ We need to apply L-Hospital's rule: So, $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\lim\limits_{n \to \infty} \dfrac{3^n \ln 3}{3n^2}=\dfrac{\infty}{\infty}$ We need to apply L-Hospital's rule again: $\lim\limits_{n \to \infty} \dfrac{3^n (\ln 3)^3}{6}=\infty$ Hence, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
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