Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 557: 63


The series $\Sigma_{n=3}^{\infty} \frac{1}{n(\ln n-1)}$ diverges.

Work Step by Step

We use the integral test as follows $$\int_3^{\infty}\frac{dn}{n(\ln n-1)}=\int_3^{\infty}\frac{d \ln n}{ (\ln n-1)}\\ =\ln(\ln n-1)|_3^{\infty}=\infty$$ Hence, the series $\Sigma_{n=3}^{\infty} \frac{1}{n(\ln n-1)}$ diverges.
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