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

Question

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

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

Answer 1

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

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

Answer 2

...and a very fancy proof: $$\,a_n:=\frac{2^n}{n!}\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\frac{2}{n+1}\xrightarrow [n\to\infty]{} 0$$

so by d'Alembert's Test, the positive series $\,\displaystyle{\sum_{n=1}^\infty\frac{2^n}{n!}}\,$ converges, and thus $$\lim_{n\to\infty}\frac{2^n}{n!}=0\Longrightarrow \lim_{n\to\infty}\frac{n!}{2^n}=\infty$$

Answer 3

HINT $$\dfrac{n!}{2^n} \geq \dfrac{n}4 \text{ for all } n \in \mathbb{N}$$

Answer 4

Here's another (somewhat roundabout) way to approach this. Noting that we've got $n!$ terms showing up, it suggests that the sequence is related to a Taylor series for some function.

Put $b_n=1/a_n$ for all $n\in\Bbb N$ and consider the function $$f(x)=\sum_{n=0}^\infty b_nx^n.$$ Rewriting this as $$f(x)=\sum_{n=0}^\infty\frac{(2x)^n}{n!},$$ we see that $f(x)$ is simply the Taylor series of $e^{2x}$ about $x=0$, which converges for all $x$. In particular, we have $$\sum_{n=0}^\infty b_n=f(1)=e^2,$$ so the series with non-negative terms $b_n$ converges. What can we then conclude about the sequence of terms $b_n$? What does that tell us about the sequence of terms $a_n=1/b_n$?