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: 25


Converges Conditionally

Work Step by Step

Consider $a_n=(-1)^{n+1}\dfrac{1+n}{n^2}$ $\Sigma_{n=1}^\infty |u_n|=\Sigma_{n=1}^\infty \dfrac{1+n}{n^2}=\Sigma_{n=1}^\infty \dfrac{1}{n^2}+\Sigma_{n=1}^\infty \dfrac{1}{n}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ a convergent p-series and $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent p-series. Thus, the sum of both series is divergent. Next, consider $b_n=\dfrac{1}{n^2}+\dfrac{1}{n}$ we can see that $b_n$ is decreasing and $\lim\limits_{n \to \infty} b_n=\lim\limits_{n \to \infty} \dfrac{1}{n^2}+\lim\limits_{n \to \infty} \dfrac{1}{n}=0$ Hence, the series Converges Conditionally by the Alternating Series Test.
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