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: 14


Sequence converges to $1$.

Work Step by Step

As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n= (\dfrac{3}{n})^{1/n}$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{3}{n})^{1/n}$ Let us consider $y=\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{3}{n})^{1/n}$ Use logarithmic rule: $\ln a^n=n \ln a$ $\ln y=\lim\limits_{n \to \infty} \dfrac{1}{n}\ln \dfrac{3}{n}=\lim\limits_{n \to \infty} \dfrac{\ln 3-\ln n}{n}$ Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, so take the help of L-Hospital's rule. $\ln y=\lim\limits_{n \to \infty} (\dfrac{-1}{n})(\dfrac{1}{1})$ $\ln y=0$ $e^{\ln y}=e^0 \implies y=1$ Therefore, the sequence converges to $1$.
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