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 487: 28

Answer

$\lim\limits_{n \to \infty} a_n=1 $ and {$a_n$} is convergent.

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n+(-1)^n}{n}$ or, $=\lim\limits_{n \to \infty} [\dfrac{n}{n}+\dfrac{(-1)^n}{n})$ or, $=\lim\limits_{n \to \infty} 1 +\lim\limits_{n \to \infty} \dfrac{(-1)^n}{n}$ or, $=1+0$ Thus, $\lim\limits_{n \to \infty} a_n=1 $ and {$a_n$} is convergent.
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