Answer
Converges
Work Step by Step
Let ${a_k} = \dfrac{1}{{{{\rm{e}}^k}}}$.
Condition 1. Examine if ${a_k} \gt {a_{k + 1}}$.
For $k \ge 1$, we have $\dfrac{1}{{{{\rm{e}}^k}}} \gt \dfrac{1}{{{{\rm{e}}^{k + 1}}}}$. Thus,
${a_k} \gt {a_{k + 1}}$
Condition 2. Examine if $\mathop {\lim }\limits_{k \to \infty } {a_k} = 0$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{{{\rm{e}}^k}}} = 0$
The two conditions above are satisfied, therefore, by Theorem 9.6.1, the alternating series converges.
