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