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: 27


$\lim\limits_{n \to \infty} a_n=2 $ and {$a_n$} is convergent.

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}[2+(0.1)^n]=\lim\limits_{n \to \infty} (2)+\lim\limits_{n \to \infty} (0.1)^n$ or, $=2+0$ or, $=2$ Thus, $\lim\limits_{n \to \infty} a_n=2 $ and {$a_n$} is convergent.
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