Answer
$=\dfrac {1}{500}=0.002$
Work Step by Step
$\sum ^{\infty }_{k=4}\dfrac {1}{5^{k}}=\dfrac {a_{1}}{1-r};a_{1}=\dfrac {1}{5^{4}};r=\dfrac {1}{5}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{5^{4}}}{1-\dfrac {1}{5}}=\dfrac {1}{500}=0.002$
