Answer
$-\dfrac {1}{7}$
Work Step by Step
$\sum ^{\infty }_{i=1}\left( -\dfrac {1}{5}\right) \left( -\dfrac {2}{5}\right) ^{i-1}=\dfrac {a_{1}}{1-r};a_{1}=\left( -\dfrac {1}{5}\right) \times \left( -\dfrac {2}{5}\right) ^{1-1}=-\dfrac {1}{5};r=-\dfrac {2}{5}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {-\dfrac {1}{5}}{1-\left( -\dfrac {2}{5}\right) }=-\dfrac {1}{7}$
