Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 28


$$ \frac{4}{5}$$

Work Step by Step

Given $$1-\frac{1}{4}+\frac{1}{4^{2}}-\frac{1}{4^{3}}+\cdots$$ Since $$1-\frac{1}{4}+\frac{1}{4^{2}}-\frac{1}{4^{3}}+\cdots=\sum_{n=0}^{\infty} \left(\frac{-1}{4}\right)^n $$ This is a geometric series with $|r|= \frac{1}{4}<1$, so the series converges and has the sum \begin{align*} S&=\frac{a}{1-r}\\ &=\frac{1}{1+\frac{1}{4}}\\ &= \frac{4}{5} \end{align*}
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