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: 47

Answer

Conditionally Convergent

Work Step by Step

A series $ \Sigma a_n$ is said to be absolutely convergent when $ \Sigma |a_n|$ is convergent. We notice that $ \Sigma_{n=2}^{\infty} |\dfrac{(-1)^n}{2n}|=\dfrac{1}{2} \Sigma_{n=2}^{\infty} \dfrac{1}{n}$ We see a p-series with common ratio $r=1$. When the common ratio $r \gt 1$, then a p-series is convergent. Therefore, the given series is not Absolutely Convergent . The Alternating Series Test states: Consider a series $\Sigma a_n$ such that $A_n=(-1)^n B_n$; $B_n \geq 0$ for all $n$ If the following conditions are satisfied, then the series converges: a) $\lim\limits_{n \to \infty} B_n=0$; b) $B_n$ is a decreasing sequence. We notice that $B_n=\dfrac{1}{2n}$ a) $\lim\limits_{n \to \infty} A_n=\lim\limits_{n \to \infty}\dfrac{1}{2n}=0$ b) We notice that the series is a decreasing sequence. Therefore, the given series is Conditionally Convergent by the Alternating Series Test.
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