Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.1 Exercises - Page 724: 50

Answer

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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