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: 20


The series $\sum_{n=2}^{\infty}\frac{ n}{n^2-n}$ diverges.

Work Step by Step

Using the limit comparison test with the divergent p-series (also known as the harmonic series) $\sum_{n=1}^{\infty} \frac{1}{n}$ Since we have $$L=\lim_{n\to \infty } \frac{n/(n^2-n)}{1/n}=\lim_{n\to \infty } \frac{ n^2}{n^2-n}\\ =\lim_{n\to \infty } \frac{ 1}{1-1/n}=1\gt0$$ then the series $\sum_{n=2}^{\infty}\frac{ n}{n^2-n}$ diverges.
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