University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 23



Work Step by Step

Consider $f(x)=\dfrac{-2}{n+1}$ or, $f(x)=\dfrac{-2}{n+1}=-2(\dfrac{1}{n+1})$ Now, take the integral test to find the convergence and divergence for the sequence. We have $\lim\limits_{x \to \infty} \int_0^\infty \dfrac{1}{ x+1}dx= \lim\limits_{a \to \infty} [\log (x+1)]_{0}^\infty=\infty$ Hence, the sequence $\Sigma_{n=0}^\infty \dfrac{-2}{ n+1}$ is Divergent.
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