University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 73


Converges to $0$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}$ Since, $\lim\limits_{n \to \infty} \dfrac{x^n}{n!}=0$ So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{3^n 6^n}{2^{-n} n!}=\lim\limits_{n \to \infty} \dfrac{(3 \cdot 2 \cdot 3 \cdot 2)^n}{n!}=\dfrac{36^n}{n!}$ or, $=0$ Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent and converges to $0$
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