Answer
$=\dfrac {5}{2}=2.5$
Work Step by Step
$\sum ^{\infty }_{k=0}\left( \dfrac {3}{5}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( \dfrac {3}{5}\right) ^{0}=1;r=\dfrac {3}{5}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {3}{5}}=\dfrac {5}{2}=2.5$
