Answer
$\dfrac {1}{2}$
Work Step by Step
$\sum ^{\infty }_{n=1}3^{-k}=\dfrac {a_{1}}{1-r};a_{1}=3^{-1}=\dfrac {1}{3};r=\dfrac {1}{3}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{3}}{1-\dfrac {1}{3}}=\dfrac {1}{2}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 11 - Further Topics in Algebra - 11.3 Geometric Sequences and Series - 11.3 Exercises - Page 1035: 59
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