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: 12



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 $p\leq 1$. $a_{n}-\frac{1}{n}=\frac{n^{2}+1}{n^{3}+1}-\frac{1}{n}$ $=\frac{(n^{2}+1)n-(n^{3}+1)}{(n^{3}+1)n}$ $=\frac{n-1}{(n^{3}+1)}\geq 0$ for all $n$ Since, $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent, $\sum_{n=1}^{\infty}\frac{n^{2}+1}{n^{3}+1}$ is also divergent. Hence, the series $\sum_{n=1}^{\infty}\frac{n^{2}+1}{n^{3}+1}$ is divergent.
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