Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 26

Answer

Divergent

Work Step by Step

Given: $\Sigma_{n =2}^{\infty} \frac{(-1)^{n}\sqrt n}{ln n}$ Check the derivative of the corresponding function. $f(x)=\frac{\sqrt x}{ln x}$ $f'(x)=\frac{lnx-2}{2\sqrt x (lnx)^{2}}$ Find where $f'(x)=0$ $ln(x)-2=0$ $ln(x)=2$ $x=e^{2}\approx 7.4$ Test the interval after the critical number. $(e^{2},\infty )$: $f'(8)\approx 0.003\gt 0$ It is positive as $x \to \infty$, so $f$ is decreasing for $x\gt e^{2}$. Since, $\frac{\sqrt n}{ln n}$ is increasing. $\lim\limits_{n \to \infty}\frac{(-1)^{n}\sqrt n}{ln n}=DNE$ The limit does not exist so the series diverges by the Test of Divergence.
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.