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: 24



Work Step by Step

Consider $f(x)=\dfrac{1}{2x-1}$ This is positive, continuous and decreasing for $x \geq 0$ Now, take the integral test to find the convergence and divergence for the sequence. We have $\lim\limits_{a \to \infty} \int_1^a \dfrac{1}{ 2x-1}dx= \lim\limits_{a \to \infty}\dfrac{1}{2} [\ln (2a-1)]=\infty$ Hence, the sequence $\Sigma_{n=0}^\infty \dfrac{1}{ 2n-1}$ is Divergent.
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