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: 31



Work Step by Step

The $n$-th Term Test states that when a series $a_n \to 0$ then the series diverges. We notice that $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (-1)^n \dfrac{n}{n+1}$ a) When $n$ is even, then $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{n}{n+1}=\lim\limits_{n \to \infty} \dfrac{n}{1+1/n}=\dfrac{1}{1+0}=1$ b) When $n$ is odd, then $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} -\dfrac{n}{n+1}=\lim\limits_{n \to \infty} -\dfrac{n}{1+1/n}=-\dfrac{1}{1+0}=-1$ Therefore, the given series Diverges by the $n$-th Term Test.
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