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: 12


Sequence converges to ${\dfrac{1}{e}}$.

Work Step by Step

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