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