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: 39

Answer

Convergent

Work Step by Step

Since, we have: $\int_1^\infty\sec h x dx=2\lim\limits_{a \to \infty} \int_1^a\frac{e^x}{1+(e^x)^2} dx$ or, $=2 [\lim\limits_{a \to \infty} [tan^{-1}(e^x)]|_{1}^{a}$ or, $ =2 \lim\limits_{a \to \infty}(\tan^{-1} e^a-tan^{-1} e)$ or, $=\pi-2 tan^{-1} e$ or, $\approx 0.71$ Hence, the series is Convergent
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.