calculus - Does the sequence $frac{n!}{2^n}$ converge or diverge?

Does the following sequence $\{a_n\}$ converge or diverge?

$$a_n=\dfrac{n!}{2^n}$$

Answer

Consider writing "out" the sequence:

$$\tag 1\frac{{n!}}{{{2^n}}} = \frac{n}{2}\frac{{n - 1}}{2}\frac{{n - 2}}{2} \cdots \frac{4}{2}\frac{3}{2}\frac{2}{2}\frac{1}{2}$$

Note that every time we take another step in the sequence, we multiply by $\dfrac{n}{2}$ so we're making the sequence larger and larger each time. In particular, we can see that every term in the factorization in $(1)$ is larger or equal than $1$, except $1/2$, so that

$$\frac{{n!}}{{{2^n}}} = \frac{n}{2}\frac{{n - 1}}{2}\frac{{n - 2}}{2} \cdots \frac{4}{2}\frac{3}{2}\frac{2}{2}\frac{1}{2} \geqslant \frac{1}{2}\frac{n}{2}=\frac{n}{4}$$

What does this tell you about the limit of the sequence?

Another approach would be D'Alambert's criterion (ratio test), which gives:

$${a_n} = \frac{{n!}}{{{2^n}}}$$

So

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)!}}{{{2^{n + 1}}}}\frac{{{2^n}}}{{n!}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{2}\frac{{n!}}{{n!}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{2} $$

What can you say about that limit? Then, what does this tell you about

$$\mathop {\lim }\limits_{n \to \infty } {a_n} \; \;?$$