Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 569: 28

Answer

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

Work Step by Step

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