Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 569: 34


$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.

Work Step by Step

Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}$ and $\lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}=\dfrac{\infty}{\infty}$ ($\bf{Indeterminate form}$) Now, apply L-Hospital's rule. $\lim\limits_{n \to \infty} \dfrac{3n^2}{8n}=\dfrac{\infty}{\infty}$ ($\bf{Indeterminate form}$) Need to apply L-Hospital's rule again. This implies , we get, $ \lim\limits_{n \to \infty} \dfrac{6n}{8}=\infty$ Hence, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
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