University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Practice Exercises - Page 552: 3

Answer

Sequence converges to $-1$.

Work Step by Step

As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=\dfrac{1-2^n}{2^ n}$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}[\dfrac{1-2^n}{2^ n}]$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{1}{2^ n}-\lim\limits_{n \to \infty}\dfrac{2^n}{2^ n}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{1}{2^ n}-1$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{1}{2})^ n-1$ $\lim\limits_{n \to \infty}a_n=0-1=-1$ Therefore, the sequence converges to $-1$.
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