Answer
Convergent
Work Step by Step
Use Limit Comparison Test with $a_{n}=(1+\frac{1}{n})^{2}e^{-n}$ and $b_{n}=e^{-n}$
$\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{(1+\frac{1}{n})^{2}e^{-n}}{e^{-n}}$
$=\lim\limits_{n \to \infty}{(1+\frac{1}{n})^{2}}$
$=(1+0)^{2}\ne 1 \ne 0 \ne \infty $
$\Sigma_{n=1}^{\infty}e^{-n}$ is convergent by the Limit Comparison Test and $\Sigma e^{-n}$ converges because it is a geometric series with $|r|\lt 1$.
