Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 46

Answer

Diverges

Work Step by Step

Given $$\sum_{n=1}^{n=\infty} \frac{1}{n-\ln n}$$ Compare with divergent series $\displaystyle\sum_{n=1}^{n=\infty} \frac{1}{n}$ ($p-$series $p=1$), by using the Limit Comparison Test \begin{align*} \lim_{n\to\infty} \frac{a_n}{b_n} &=\lim_{n\to\infty} \frac{ \frac{1}{n-\ln n}}{\frac{1}{n}}\\ &=\lim_{n\to \infty} \frac{n}{n-\ln n}\\ &=\lim_{n\to \infty} \frac{1}{1-\frac{1}{n} }\\ &=1 \end{align*} Then the given series also diverges.
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