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: 31



Work Step by Step

As per the $n$-th Term Test, when a series $a_n \to 0$ , then the series diverges. We have : $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (-1)^n \dfrac{n}{n+1}$ (a) Consider that $n$ is even; then we have : $\lim\limits_{n \to \infty} a_n =\lim\limits_{n \to \infty} \dfrac{n}{n+1}\\=\lim\limits_{n \to \infty} \dfrac{n}{1+(1/n)} \\=1$ b) Consider that $n$ is odd; then we have: $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} -\dfrac{n}{n+1} \\=\lim\limits_{n \to \infty} -\dfrac{n}{1+(1/n)} \\=-1$ This implies that the series Diverges by the $n$-th Term Test.
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