#### Answer

$\dfrac {1}{2}$

#### Work Step by Step

$\sum ^{\infty }_{n=1}3^{-k}=\dfrac {a_{1}}{1-r};a_{1}=3^{-1}=\dfrac {1}{3};r=\dfrac {1}{3}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{3}}{1-\dfrac {1}{3}}=\dfrac {1}{2}$

Published by
Pearson

ISBN 10:
013421742X

ISBN 13:
978-0-13421-742-0

$\dfrac {1}{2}$

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