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 10.3 - The Integral Test - Exercises 10.3 - Page 586: 23



Work Step by Step

Let us consider $f(x)=\dfrac{-2}{n+1}$ Re-write as: $f(x)=-2(\dfrac{1}{n+1})$ Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 0$ Then $\lim\limits_{x \to \infty} \int_0^\infty \dfrac{1}{ x+1}dx= \lim\limits_{k \to \infty} [\log (x+1)]_{0}^k=\ln (k+1)-\ln 1=\infty$ Thus, the sequence $\Sigma_{n=0}^\infty \dfrac{-2}{ n+1}$ is Divergent.
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