Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises - Page 778: 31

Answer

Divergent

Work Step by Step

By Root Test, we have $\lim_{n\to\infty} \sqrt[n]{ \biggl|\frac{n}{\ln n}\biggr|^n}=\lim_{n\to\infty}\frac{n}{\ln n} $ We can throw the absolute sign since $n \geq 2\implies\frac{n}{\ln n} > 0$ And since the limit is indeterminate form $\frac{\infty}{\infty}$, we can apply L'Hopital's Rule, $\lim_{x\to\infty}\frac{x}{\ln x}=\lim_{x\to\infty}x=\infty$ The above equation implies $\lim_{n\to\infty}\frac{n}{\ln n } = \infty$ Since, the limit is $> 1$, the series is divergent
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