Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.1 An Overview - 8.1 Exercises - Page 605: 38

Answer

$a_1 = \frac{2}{3}$ $a_2 =\frac{2}{3}$ $a_3 = \frac{2}{3}$ $a_4 = \frac{2}{3}$ Converges. The limit of the sequence approaches $\frac{2}{3}$.

Work Step by Step

$a_{n+1} = 1-\frac{a_n}{2}$; $a_0 = \frac{2}{3}$ $a_1 = 1- \frac{a_0}{2} = \frac{2}{3}$ $a_2 = 1- \frac{a_1}{2} =\frac{2}{3}$ $a_3 = 1- \frac{a_2}{2} = \frac{2}{3}$ $a_4 = 1- \frac{a_3}{2} = \frac{2}{3}$ Converges and approaches $\frac{2}{3}$. The terms of the sequence are constant.
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