Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 537: 19



Work Step by Step

Since we have $$\lim_{n\to \infty}c_n=\lim_{n\to \infty}-2^{-n}=\lim_{n\to \infty}-\frac{1}{2^n}=0$$ then the sequence $c_n$ converges to $0$.
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