Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 22



Work Step by Step

The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. $a_{n}=\frac{\sqrt {n+1}-\sqrt {n-1}}{n}$ $=\frac{(\sqrt {n+1}-\sqrt {n-1})(\sqrt {n+1}+\sqrt {n-1})}{n(\sqrt {n+1}+\sqrt {n-1})}$ $=\frac{2}{n(\sqrt {n+1}+\sqrt {n-1})} \lt \frac{2}{n\sqrt {n+1}} \lt \frac{2}{n\sqrt {n}}=\frac{2}{n^{3/2}}$ Thus, the series convergent.
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