Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 27



Work Step by Step

Given $$ \sum_{n=-4}^{n=\infty} \left(\frac{-4}{9}\right)^{n}$$ Since the series is a geometric series with $|r|=\left|\frac{-4}{9}\right|<1$ , then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{ \left(\frac{-4}{9}\right)^{-4} }{1-\frac{-4}{9}}\\ &=\frac{59049}{3328} \end{align*}
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