Answer
$=\dfrac {9}{460}\approx 0.019565$
Work Step by Step
$\sum ^{\infty }_{k=2}\left( -0.15\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( -0.15\right) ^{2}=0.0225;r=\left( -0.15\right) \Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {0.0225}{1-\left( -0.15\right) }=\dfrac {225}{11500}=\dfrac {9}{460}\approx 0.019565$