Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 736: 34


The series converges for all the values of $p$ (can be any real number).

Work Step by Step

Case 1: When $p\leq 0$ $\lim\limits_{n \to \infty}\dfrac{(\ln n)^p}{n}=0$ . Thus, the series converges by the Test of Divergence. Case 2: When $p \gt 0$ $\lim\limits_{n \to \infty}\dfrac{(\ln n)^p}{n}=\lim\limits_{n \to \infty}\dfrac{p(\ln n)^{p-1}(1/n)}{1}$ or, $=\lim\limits_{n \to \infty}\dfrac{p(\ln n)^{p-1}}{n}$ Case 3: When $p \leq 1$ $\lim\limits_{n \to \infty}\dfrac{(\ln n)^p}{n}=\lim\limits_{n \to \infty}\dfrac{p(\ln n)^{p-1}(\dfrac{1}{n})}{1}$ and $\lim\limits_{n \to \infty}\dfrac{p(\ln n)^{p-1}}{n}=0$ This implies that the limit $0$ satisfies all the conditions for the alternating series test and so, the series converges by the Test of Divergence. Hence, the series will converge for all values of $p$ (any real number).
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