Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.6 - Alternating Series and Conditional Convergence - Exercises 10.6 - Page 603: 28

Answer

Converges conditionally

Work Step by Step

Alternating Series Test states that consider a series $\Sigma a_n$ such that $p_n=(-1)^n q_n$; $q_n \geq 0$ for all $n$ When the following conditions are satisfied then the series converges such that i) $\lim\limits_{n \to \infty} q_n=0$; ii) $q_n$ is a decreasing sequence. Here, we have $q_n=\dfrac{1}{n \ln n}$ i) Now, $\lim\limits_{n \to \infty} q_n=\lim\limits_{n \to \infty}\dfrac{1}{(n)(\ln n)}=\dfrac{1}{\infty}=0$; ii) Also, $q_n=\dfrac{1}{n \ln n}$ shows a decreasing sequence as the function $n \ln n$ increases. Hence, the given series Converges conditionally by the Alternating Series Test.
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