Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 13

Answer

Divergent

Work Step by Step

$\frac{1}{3}+\frac{1}{7}+\frac{1}{11}+\frac{1}{15}+\frac{1}{19}+...=\Sigma \frac{1}{4n-1}$ $\int_{1}^{\infty}\frac{1}{4x-1}dx=\lim\limits_{t \to \infty}\int_{1}^{t}\frac{1}{4x-1}dx=\lim\limits_{t \to \infty}[\frac{1}{4}ln(4x-1)]_{1}^{t}=\infty-\frac{1}{4}ln(3)=\infty$ Divergent
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