Answer
The series diverges.
Work Step by Step
Let ${a_k} = {\left( {1 + \dfrac{1}{k}} \right)^{ - k}}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^{ - k}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{{{\left( {1 + \dfrac{1}{k}} \right)}^k}}} = \dfrac{1}{{\mathop {\lim }\limits_{k \to \infty } {{\left( {1 + \dfrac{1}{k}} \right)}^k}}}$
It is known (see Eq. 6 in Section 6.1) that $\mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^k} = {\rm{e}}$. Thus,
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \dfrac{1}{{\rm{e}}}$
Since $\mathop {\lim }\limits_{k \to \infty } {a_k} \ne 0$, by the divergence test, the series diverges.