Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 32


The series $\Sigma_{n=2}^{\infty} \frac{1}{(\ln n)^2}$ diverges.

Work Step by Step

Using the property that $\ln n \lt n^{1/2}$, then $(\ln n)^2\lt n$ and hence we have $$\frac{1}{(\ln n)^2}\geq \frac{ 1}{n }$$ Since $\Sigma_{n=2}^{\infty}\frac{ 1}{n}$ is a divergent p-series ( $p=1$) then the series $\Sigma_{n=2}^{\infty} \frac{1}{(\ln n)^2}$ diverges by the comparison test.
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