Answer
$=\dfrac {10}{19}$
Work Step by Step
$\sum ^{\infty }_{n=0}\left( -\dfrac {9}{10}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( -\dfrac {9}{10}\right) ^{0}=1;r=-\dfrac {9}{10}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\left( -\dfrac {9}{10}\right) }=\dfrac {10}{19}$