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 538: 53


The sequence $y_n$ diverges.

Work Step by Step

We have $$\lim_{n\to \infty}y_n=\lim_{n\to \infty} \frac{e^n}{2^n} \\ = \lim_{n\to \infty} (e/2)^n = \infty .$$ Since we used the fact that $e/2\gt 1. $ Hence, by Theorem 1, the sequence $y_n$ diverges.
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