University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.6 - Alternating Series and Conditional Convergence - Exercises - Page 521: 11

Answer

Converges

Work Step by Step

Consider $u_n=\dfrac{\ln n}{n}$ Here, $u_n$ is positive for all the values of $n$. Now, $f(n)=\dfrac{\ln n}{n}$ and $f'(n)=\dfrac{1-\ln n}{n^2} \leq 0$ when $n \geq 3$ The negative sign shows that, the sequence $u_n$ is not increasing. Thus, $\lim\limits_{n \to \infty} u_n=\lim\limits_{n \to \infty}\dfrac{\ln n}{n}$ Need to apply L-Hospital's rule. we have $\lim\limits_{n \to \infty}\dfrac{\ln n}{n}=\lim\limits_{n \to \infty}\dfrac{1/n}{1}=0$ Hence, by the Alternating Series Test, the series converges.
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