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: 35

Answer

$a_1 = \frac{(-1)^1}{2^1} = \frac{-1}{2}$ $a_2 = \frac{(-1)^2}{2^2} = \frac{1}{4}$ $a_3 = \frac{(-1)^3}{2^3} = \frac{-1}{8}$ $a_4 = \frac{(-1)^4}{2^4} = \frac{1}{16}$ Converges. The limit of the sequence seems to be $0$.

Work Step by Step

$a_n = \frac{(-1)^n}{2^n}$ $a_1 = \frac{(-1)^1}{2^1} = \frac{-1}{2}$ $a_2 = \frac{(-1)^2}{2^2} = \frac{1}{4}$ $a_3 = \frac{(-1)^3}{2^3} = \frac{-1}{8}$ $a_4 = \frac{(-1)^4}{2^4} = \frac{1}{16}$ Converges and approaches $0$. Although the signs of the terms alternate in the sequence, the absolute value of the terms is decreasing and seems to approach $0$.
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