Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 704: 41

Answer

Converges to $0$

Work Step by Step

Given: $a_n=n^{2}e^{-n}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n^{2}e^{-n}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{n^{2}}{e^{n}}$ Use L-Hospital's Rule. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{2n}{e^{n}}$ Again use L-Hospital's Rule. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{2}{e^{n}}$ $=\frac{2}{\infty}$ $=0$ Hence, the sequence converges to $0$.
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