Given number:
$$\frac{\sqrt[3]{n} n!}{3^{n} + 2}$$
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
$$a_{n} = \frac{\sqrt[3]{n} n!}{3^{n} + 2}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt[3]{n} \left(3^{n + 1} + 2\right) \left|{\frac{n!}{\left(n + 1\right)!}}\right|}{\left(3^{n} + 2\right) \sqrt[3]{n + 1}}\right)$$
Let's take the limitwe find
False
False