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: 21

Answer

Conditionally convergent

Work Step by Step

Consider $a_n=\dfrac{1}{n+3}$ Here, $a_n$ is positive for all the values of $n$. Let $f(n)=\dfrac{1}{n+3} \implies f'(n)=\dfrac{-1}{(n+3)^2}\lt 0$ The negative sign shows that, the sequence $u_n$ is not increasing. Thus, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{n+3}=0$ Thus, by the Alternating Test Series, the series converges. We know that $\Sigma_{n=1}^\infty \dfrac{1}{ n}$ diverges. Thus, the series does not converge absolutely. Hence, the series is conditionally convergent.
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