Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 50


Converges to $0$

Work Step by Step

Given:$a_n=\frac{(lnn)^{2}}{n}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{(lnn)^{2}}{n}$ Apply L-Hospital's Rule. $=\lim\limits_{n \to \infty}\frac{2ln (n).\frac{1}{n}}{1}$ $=\lim\limits_{n \to \infty}\frac{2ln (n)}{n}$ Again apply L-Hospital's Rule. $=\lim\limits_{n \to \infty}\frac{2}{n}$ $=0$ Hence, the sequence converges to $0$.
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