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: 23


The series $\sum_{n=1}^{\infty} \frac{ 1}{\sqrt{n^2+1}}$ diverges.

Work Step by Step

Use the limit comparison test with the p-series $\sum_{n=1}^{\infty} \frac{1}{n}$ Since we have $$L=\lim_{n\to \infty } \frac{1/\sqrt{n^2+1}}{1/n }=\lim_{n\to \infty } \frac{ n }{\sqrt{n^2+1}}\\ =\lim_{n\to \infty } \frac{ 1}{\sqrt{1+1/n^2}}=1\gt0$$ then the series $\sum_{n=1}^{\infty} \frac{ 1}{\sqrt{n^2+1}}$ diverges because the p-series $\sum_{n=1}^{\infty} \frac{ 1}{ n }$, $p=1$, diverges.
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