Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 11

Answer

Convergent

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 $p\leq 1$. $a_{n}=\frac{n}{n^{3}+1}\lt \frac{n}{n^{3}}=\frac{1}{n^{2}}$ Since, $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is convergent, $\sum_{n=1}^{\infty}\frac{n}{n^{3}+1}$ is also convergent. Hence, the series $\sum_{n=1}^{\infty}\frac{n}{n^{3}+1}$ is convergent.
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