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