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: 67


Converges to $e^{-1}$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} [(\dfrac{1}{n})^{1/\ln n}]$ Since, $e^{\ln x}=x$ So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} [(\dfrac{1}{n})^{1/\ln n}]= \lim\limits_{n \to \infty} e^{\frac{1}{\ln n} \ln (1/n)}= \lim\limits_{n \to \infty} e^{\frac{\ln 1 -\ln n}{\ln n}}=e^{-1}$ Hence, $\lim\limits_{n \to \infty} a_n=e^{-1}$ and {$a_n$} is Convergent and converges to $e^{-1}$
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