\displaystyle \limsup_{n \to +\infty} \frac{\frac{2^{3 n + 4}}{(n + 1)!}}{\frac{2^{3 n + 1}}{n!}} = \lim_{n \to +\infty} \frac{8}{n + 1} = 0 \ \Rightarrow \ R = +\infty