Answer
$=\dfrac {1}{7}$
Work Step by Step
$\sum ^{\infty }_{n=1}\left( 2^{-3k}\right) =\sum ^{\infty }_{k=1}8^{-k}=\dfrac {a_{1}}{1-r};a_{1}=8^{-1};r=\dfrac {1}{8}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{8}}{1-\dfrac {1}{8}}=\dfrac {1}{7}$