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: 41


Conditionally Convergent

Work Step by Step

A series $ \Sigma a_n$ is defined as absolutely convergent when the series $ \Sigma |a_n|$ is convergent. From the series, we notice that $ \Sigma_{n=1}^{\infty} |(-1)^n(\sqrt {n+1}-\sqrt n)|= \Sigma_{n=1}^{\infty} (\sqrt {n+1}-\sqrt n)=\Sigma_{n=2}^{\infty} \dfrac{1}{n^{1/2}}$ This corresponds to a p-series with common ratio $r=\dfrac{1}{2} \lt 1$. Remember that a p-series is said to be convergent when the common ratio $ r \gt 1$. This implies that the series is Not Absolutely Convergent. The Alternating Series Test states that a series $\Sigma a_n$ such that $p_n=(-1)^n q_n$; $q_n \geq 0$ for all $n$ If the following conditions are satisfied then the series converges: a) $\lim\limits_{n \to \infty} q_n=0$; b) $q_n$ is a decreasing sequence. We have: $q_n=\dfrac{1}{(\sqrt {n+1}-\sqrt n)}$ a) $\lim\limits_{n \to \infty} q_n=\lim\limits_{n \to \infty}\dfrac{1}{(\sqrt {n+1}-\sqrt n)}=0$ b) The series is a decreasing sequence. This implies that the given series is Conditionally Convergent by the Alternating Series Test.
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