Answer
$\dfrac {-8}{15}$
Work Step by Step
$\sum ^{\infty }_{i=1}\left( -\dfrac {2}{3}\right) \left( -\dfrac {1}{4}\right) ^{i-1}=\dfrac {a_{1}}{1-r};a_{1}=\left( -\dfrac {2}{3}\right) \times \left( -\dfrac {1}{4}\right) ^{1-1}=-\dfrac {2}{3};r=\left( -\dfrac {1}{4}\right) \Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {-\dfrac {2}{3}}{1-\left( -\dfrac {1}{4}\right) }=\dfrac {-8}{15}$