Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 7



Work Step by Step

Given: $a_{n}=(1+\frac{3}{n})^{4n}$ We know that $\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n}=e$ Assume that $\frac{n}{3}x=4n$ and solve for $x$. we get $x=12$ and $\frac{n}{3}\times12=4n$ Thus, $\lim\limits_{n \to \infty}(1+\frac{3}{n})^{4n}=\lim\limits_{n \to \infty}[(1+\frac{3}{n})^{n/3}]^{12}$ Therefore, $\lim\limits_{n \to \infty}[(1+\frac{3}{n})^{n/3}]^{12}=e^{12}$ Because the limits exists , the sequence converges. Hence, the given sequence converges to $e^{12}$.
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