Answer
Converges
Work Step by Step
We use the Limit Comparison Test with
$a_n=(1+\frac{1}{n})^2e^{-n}$ and $b_n=e^{-n}$
and obtain
$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=\lim\limits_{n \to \infty}\frac{(1+\frac{1}{n})^2e^{-n}}{e^{-n}}=\lim\limits_{n \to \infty}(1+\frac{1}{n})^2=(1+0)^2=1>0$.
Notice that $\sum_{n=1}^\infty e^{-n}=\sum_{n=1}^\infty (\frac{1}{e})^n$ is a geometric series with the common ratio $1/e$.
