## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.4 The Comparison Tests - 11.4 Exercises - Page 771: 19

convergent

#### Work Step by Step

Given $$\sum_{n=1}^{\infty} \frac{ n+1 }{n^3+n}$$ Use the Limit Comparison Test with $a_{n}=\dfrac{ n+1 }{n^3+n}$ and $b_{n}=\dfrac{1}{ n^2}$ \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{ n^3+n^2 }{n^3+n}\\ &=1 \end{align*} since $\displaystyle\sum_{n=1}^{\infty} \frac{1}{ n^2}$ is convergent $(p-\text { series } p>1),$ then $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n^{3}+1}$ also convergent

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