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: 30

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 $n$ The series converges when the following conditions are satisfied such as follows: i) $\lim\limits_{n \to \infty} q_n=0$; ii) $q_n$ is a decreasing sequence. Here, $q_n=\dfrac{\ln n}{n- \ln n}$ i) $\lim\limits_{n \to \infty} p_n=\lim\limits_{n \to \infty}\dfrac{\ln n}{n- \ln n}=\dfrac{1/n}{1-1/n}=0$; ii) Here, $q_n=\dfrac{\ln n}{n- \ln n}$ shows a decreasing sequence we can see that $q'_n=\dfrac{1-\ln n}{(n- \ln n)^2}\lt 0$ Hence, the given series Converges conditionally by the Alternating Series Test.
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