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: 3



Work Step by Step

A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant. $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{n^{3}}{1+n^{2}}$ Divide numerator and denominator by $n^{3}$. $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{\frac{n^{3}}{n^{3}}}{\frac{1+n^{2}}{n^{3}}}$ $=\lim\limits_{n \to \infty}\frac{1}{\frac{1}{n^{3}}+\frac{1}{n}}$ $=\frac{1}{0+0}$ $=\infty$ We can also use L-Hospital's rule because the limit is of the form $\frac{\infty}{\infty}$ After applying L-Hospital's rule we get $\lim\limits_{n \to\infty}\frac{3n^{2}}{2n}=\lim\limits_{n \to\infty}\frac{3n}{2}=\infty$ Hence, the given sequence is divergent.
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