Given number:
$$\frac{\left(-1\right)^{n} x^{2 n}}{\left(2 n\right)!}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \frac{\left(-1\right)^{n}}{\left(2 n\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 2$$
,
$$c = 1$$
then
$$R^{2} = \lim_{n \to \infty} \left|{\frac{\left(2 n + 2\right)!}{\left(2 n\right)!}}\right|$$
Let's take the limitwe find
$$R^{2} = \infty$$
$$R = \infty$$