## Calculus: Early Transcendentals 8th Edition

Suppose that we have series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$. Then if the following two condition are satisfied the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. Given: $\Sigma_{n=1}^{\infty}{(-1)^{n}}{e^{-n}}$ In the given problem, $b_{n}={e^{-n}}=\frac{1}{{e^{n}}}$ which satisfies both conditions of Alternating Series Test as follows: 1. $b_{n+1}\leq\ b_{n}$ $\frac{1}{{e^{n+1}}}\leq \frac{1}{{e^{n}}}$is true. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{1}{{e^{n}}}=\frac{1}{\infty}=0$ Hence, the given series is convergent by Alternating Series Test.