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 488: 83


Converges to $0$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (\dfrac{1}{3})^n+\dfrac{1}{\sqrt 2^n}$ Since, $ \lim\limits_{n \to \infty} x^n=0$ So, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (\dfrac{1}{3})^n+\dfrac{1}{\sqrt 2^n}=\lim\limits_{n \to \infty} [(\dfrac{1}{3})^n+(\dfrac{1}{\sqrt 2})^n]=0$ Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is Convergent and converges to $0$
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