## Calculus: Early Transcendentals 8th Edition

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 $|r|\lt 1$.