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: 34

Answer

$0$

Work Step by Step

Definition of a sequence defined by a function: Let us consider a function $f(x)$ whose limit $\lim\limits_{x \to \infty} f(x)$ exists then the sequence $a_n=f(n)$ will converge to the same limit. That is, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$ Here, we have $a_n=\dfrac{(-1)^{n+1} \ n^2}{2n^3+n}$ and $f(x)=\dfrac{(-1)^{x+1} \ x^2}{2x^3+x}$ Next, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{(-1)^{x+1} \ x^2}{2x^3+x}=\lim\limits_{x \to \infty} \dfrac{(-1)^{x+1} (1/x)}{(2+\dfrac{1}{x^2})}$ a) When $n$ is odd and $n+1$ is even. $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{-1 (1/x)}{(2+\dfrac{1}{x^2})}=\dfrac{-1(0)}{2+0}=0$ b) When $n$ is even and $n+1$ is odd. $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{1 (1/x)}{(2+\dfrac{1}{x^2})}=\dfrac{(1)(0)}{2+0}=0$ Thus,$\lim\limits_{n \to \infty} a_n=0$
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