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: 32

Answer

Convergent

Work Step by Step

Use u-substitution method. Plug $u=\ln x $ and $ du=\dfrac{1}{x} dx$ Now, $\int_1^\infty \dfrac{\dfrac{1}{x}}{ \sqrt {1+(\ln^2 x)}} dx=\int_{0}^\infty \dfrac{1}{(1+u^2)} du$ This implies that $\lim\limits_{k \to \infty} [\tan^{-1} (u)]_{0}^k=\lim\limits_{k \to \infty}[(\tan^{-1} (k) -\tan^{-1} (0)]$ so, $\dfrac{\pi}{2}-0=\dfrac{\pi}{2}$ Thus, 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.