Limit is Haunting! (3)

For $n \ge 3$ we have that $\dfrac{2^{n}}{n!} = \dfrac{2 \times 2 \times 2 \times 2 \times ..... \times 2}{1 \times 2 \times 3 \times 4 \times ..... \times n} \le \dfrac{2}{1} \times \dfrac{2}{2} \times \left(\dfrac{2}{3}\right)^{n - 2}$ ,

which goes to $\boxed{0}$ as $n \to \infty$ since $\dfrac{2}{3} \lt 1$ .

Comment: Note that $\displaystyle\sum_{n=0}^{\infty} \dfrac{2^{n}}{n!} = e^{2}$ , which couldn't converge unless it was the case that $\displaystyle \lim_{n \to \infty} \dfrac{2^{n}}{n!} = 0$ ,

(a necessary but not sufficient condition for convergence of a series; the ratio test guarantees convergence in this case).