Prove that $\lim\limits_{n\to\infty}\frac{2^n}{n!}=0,n\in \mathbb{N}$

Question

If we know that $\frac{2^n}{n!}>0$ for every $n\in \mathbb{N}$ and $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}...\frac{2}{n}$$ how to bound this sequence above?

Answer 1

You cannot know that $\lim_{n\to +\infty}\frac{2^n}{n!}\color{red}{>}0$ since the limit is exactly zero.

That follows from the trivial inequality: $$ \forall n\geq 3,\qquad \frac{2^n}{n!}\leq \frac{2^n}{2\cdot 3^{n-2}}=\frac{9}{2}\cdot\left(\frac{2}{3}\right)^n\xrightarrow[n\to +\infty]{}0.$$

Answer 2

Hint: Stirling's approximation

Answer 3

For $n \ge 4$ you have $$\frac{2}{1} \frac{2}{2} \frac{2}{3} \cdots \frac{2}{n} \le 2 \cdot 1 \cdot 1 \cdot \dots 1 \cdot \frac{2}{n} = \frac{4}{n}$$

Answer 4

$$a_n:=\frac{2^n}{n!}\implies\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\frac2{n+1}\xrightarrow[n\to\infty]{}0$$

so by the ratio test (d'Alembert's), we get that the series

$$\sum_{n=1}^\infty\frac{2^n}{n!}\;\;\;\text{converges}\;\;\;\implies\;\;\;\lim_{n\to 0}\frac{2^n}{n!}=0$$