Answer
$4$
Work Step by Step
$\sum ^{\infty }_{n=0}\left( \dfrac {4}{3}\right) ^{-n}=\sum ^{\infty }_{k=0}\left( \dfrac {3}{4}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( \dfrac {3}{4}\right) ^{0}=1;r=\dfrac {3}{4}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {3}{4}}=4$
