Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Sections 11.1-11.3 - Integrated Review - Sequences and Series - Page 653: 7

Answer

$a_{40}=\dfrac{1}{40}$

Work Step by Step

Substituting $n=40$ in the expression $\dfrac{(-1)^n}{n}$, then, \begin{array}{l} a_{40}=\dfrac{(-1)^{40}}{40}\\\\ a_{40}=\dfrac{1}{40} .\end{array}
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