$z \in \mathbb C$ here.
It's a power series with coefficints $c_n := \frac{(n!)^2}{(2n)!}$, hence part of the work is the computation of the convergence radius. Here it is: $$ \frac{c_{n+1}}{c_n} = \frac{(n+1)^2}{(2n+1)(2n+2)} \to \frac14 \quad \text{as } n \to +\infty$$ That is, the radius is $4$ and the series converges for $\lvert z \rvert < 4$.
What if $\lvert z \rvert = 4$? If $z = 4$, then the series does not converge. This is done by using the Stirling formula [it is suggested me to do so]: the terms are $$ c_n 4^n \sim \frac{2 \pi n \left( \frac{n}{e} \right)^{2n}}{\sqrt{2 \pi 2 n} \left( \frac{2n}{e} \right)^{2n}} 4^n = \sqrt{\pi n} \quad \text{as } n \to +\infty$$ Consequently, the series does not converge for $z = 4$.
But what if $\lvert z \rvert = 4$ and $z \ne 4$? The sequence of the $c_n$'s is decreasing and convergent to $0$, hence we can apply the Dirichlet test, to say that in this case converges.
Is the solution right?