Answer
The Ratio Test; converges.
Work Step by Step
We apply the Ratio Test. Write ${a_k} = \dfrac{{{{\left( {k!} \right)}^2}}}{{\left( {2k} \right)!}}$.
Evaluate:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_{k + 1}}}}{{{a_k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{{\left[ {\left( {k + 1} \right)!} \right]}^2}}}{{\left[ {2\left( {k + 1} \right)} \right]!}}\cdot\dfrac{{\left( {2k} \right)!}}{{{{\left( {k!} \right)}^2}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{{\left[ {\left( {k + 1} \right)k!} \right]}^2}}}{{\left( {2k + 2} \right)!}}\cdot\dfrac{{\left( {2k} \right)!}}{{{{\left( {k!} \right)}^2}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{{\left( {k + 1} \right)}^2}}}{{\left( {2k + 1} \right)\left( {2k + 2} \right)}}$
$ = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^2} + 2k + 1}}{{4{k^2} + 6k + 2}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{1 + \dfrac{2}{k} + \dfrac{1}{{{k^2}}}}}{{4 + \dfrac{6}{k} + \dfrac{2}{{{k^2}}}}} = \dfrac{1}{4}$
Since $\rho = \dfrac{1}{4} \lt 1$, by the Ratio Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( {k!} \right)}^2}}}{{\left( {2k} \right)!}}$ converges.
