University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Practice Exercises - Page 552: 15

Answer

Converges to $\ln 2$.

Work Step by Step

As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=n(2^{1/n}-1)$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n(2^{1/n}-1)$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n2^{1/n}-\lim\limits_{n \to \infty}(1)=\lim\limits_{n \to \infty}\dfrac{2^{1/n}-1}{1/n}$ Since, we can see that the limit has the form of $\frac{0}{0}$, so take the help of L-Hospital's rule. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2^{1/n} \ln 2}{1}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2^{1/n} \ln 2}{1}=\ln 2$ Therefore, the sequence converges to $\ln 2$.
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