Answer
$=\dfrac {\pi }{\pi -e}$
Work Step by Step
$1+\dfrac {e}{\pi }+\dfrac {e^{2}}{\pi ^{2}}+\dfrac {e^{3}}{\pi ^{3}}-..=\dfrac {a_{1}}{1-r};a_{1}=1;r=\dfrac {e}{\pi }\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {e}{\pi }}=\dfrac {\pi }{\pi -e}$
