University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 72


Converges to $1$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})^n$ But $ \lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})=\lim\limits_{n \to \infty} \dfrac{\ln (1-\dfrac{1}{n^2})}{1/n}=\dfrac{0}{0}$ Need to apply L-Hospital's rule. So, $\lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})=\lim\limits_{n \to \infty} \dfrac{-2n}{n^2-1}=0$ Since, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$ when $x \gt 0$ Thus, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} e^{n \ln n (1-\dfrac{1}{n^2})}=e^0=1$ Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is convergent and converges to $1$
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