University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 497: 1


$s_n=3(1-\dfrac{1}{3^n})$ and series sum $=3$

Work Step by Step

Consider the series, $ 2+\dfrac{2}{3}+\dfrac{2}{9}+\dfrac{2}{27}+...$ The given sequence shows a geometric series with common ratio, $r=\dfrac{1}{3}$ The Nth partial sum of a geometric series can be found as: $s_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{2(1-(\dfrac{1}{3})^n)}{1-\dfrac{1}{3}}=3(1-\dfrac{1}{3^n})$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=\dfrac{2}{1-\dfrac{1}{3}}=3$ Hence, $s_n=3(1-\dfrac{1}{3^n})$ and series sum $=3$
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