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



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.
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