Answer
Converges absolutely
Work Step by Step
Let ${a_k} = {\left( { - 1} \right)^{k + 1}}\dfrac{{{2^k}}}{{k!}}$.
Evaluate the limit:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\left| {{a_{k + 1}}} \right|}}{{\left| {{a_k}} \right|}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{\dfrac{{{2^{k + 1}}}}{{\left( {k + 1} \right)!}}}}{{\dfrac{{{2^k}}}{{k!}}}} = \dfrac{{{2^{k + 1}}}}{{\left( {k + 1} \right)!}}\cdot\dfrac{{k!}}{{{2^k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{2}{{k + 1}} = 0$
Since $\rho = 0 \lt 1$, by the ratio test for absolute convergence, the series converges absolutely. Hence, it converges.
