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: 71


Converges to $x$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{x^n}{2n+1})^{1/n}$ Since, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$ when $x \gt 0$ So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{x^n}{2n+1})^{1/n}= \lim\limits_{n \to \infty} x(\dfrac{1}{2n+1})^{1/n}=x\lim\limits_{n \to \infty} e^{\dfrac{1}{n}\ln (\dfrac{1}{2n+1})}$ or, $=(x)(e^{0})$ or, $=x$ Hence, $\lim\limits_{n \to \infty} a_n=x$ and {$a_n$} is convergent and converges to $x$
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