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

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

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    See also [Show that the sequence $\left(\frac{2^n}{n!}\right)$ has a limit.](https://math.stackexchange.com/q/251441) and [Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.](https://math.stackexchange.com/q/77550).2017-04-17

4 Answers 4

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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

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

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

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    @JonasMeyer That is true. My point is that it is better either hint or explain (depending on your stance on HW questions) how to **deduce** an useful inequality, rather than just to give it out plainly. This will help him in this case only, but nourishing the deductive abilities will help in broader cases. You know: "Teach a man to fish".2012-06-28
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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 $a_n=1/b_n$?

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...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 $\lim_{n\to\infty}\frac{2^n}{n!}=0\Longrightarrow \lim_{n\to\infty}\frac{n!}{2^n}=\infty$