Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.2 Sequences - 8.2 Exercises - Page 616: 20

Answer

$\dfrac{1}{e^5}$

Work Step by Step

We are given the sequence: $\left\{\left(\dfrac{n}{n+5}\right)^n\right\}$ Rewrite the general term of the sequence: $\left(\dfrac{n}{n+5}\right)^n=\left(\dfrac{1}{\dfrac{n+5}{n}}\right)^n=\left(\dfrac{1}{1+\dfrac{5}{n}}\right)^n$ $=\dfrac{1^n}{\left(1+\dfrac{5}{n}\right)^n}=\dfrac{1}{\left(1+\dfrac{5}{n}\right)^n}=\dfrac{1}{\left(1+\dfrac{1}{\frac{n}{5}}\right)^n}$ $=\dfrac{1}{\left(\left(1+\dfrac{1}{\frac{n}{5}}\right)^{n/5}\right)^5}$ Compute the limit: $\lim\limits_{n \to \infty}\dfrac{1}{\left(\left(1+\dfrac{1}{\frac{n}{5}}\right)^{n/5}\right)^5}=\dfrac{1}{\lim\limits_{n \to \infty}\left(\left(1+\dfrac{1}{\frac{n}{5}}\right)^{n/5}\right)^5}=\dfrac{1}{\left(\lim\limits_{n \to \infty}\left(1+\dfrac{1}{\frac{n}{5}}\right)^{n/5}\right)^5}$ Because $\lim\limits_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n=e$, we have: $\dfrac{1}{\left(\lim\limits_{n \to \infty}\left(1+\dfrac{1}{\frac{n}{5}}\right)^{n/5}\right)^5}=\dfrac{1}{e^5}$
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