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 limit we find