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: 49


$\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}$ But $\lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}=\dfrac{\infty}{\infty}$ We need to apply L-Hospital's rule: So, $\lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}=\lim\limits_{n \to \infty} \dfrac{ 2\sqrt n}{n+1}$ or, $\lim\limits_{n \to \infty} \dfrac{\dfrac{2}{\sqrt n}}{1+\dfrac{1}{n}}=0$ Hence, $\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
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