Answer
The series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{2^{2n}}}}{{n!}}$ converges absolutely.
Work Step by Step
Compute the ratio and its limit with ${a_n} = \dfrac{{{2^{2n}}}}{{n!}}$.
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{2^{2n + 2}}}}{{\left( {n + 1} \right)!}}\cdot\dfrac{{n!}}{{{2^{2n}}}}} \right| = 4\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{n + 1}}} \right| = 0$
Since $\rho \lt 1$, by Ratio Test, the series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{2^{2n}}}}{{n!}}$ converges absolutely.
