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

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

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3  
Hint: try working out the values of the terms for values of $n$ like 10, 20, 30, ... then make a guess at the probable limit, and try to find a mathematical reason why it might be true? Ask us when/if you get stuck. –  Old John Jun 27 '12 at 19:44
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@OldJohn: I am a lecturer at a collage and suggest the same way to my students who don't know what to do at first. :) –  Babak S. Jun 27 '12 at 19:48

4 Answers 4

up vote 7 down vote accepted

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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Part of Peter's subsequent answer expands on this hint. Perhaps user1405177 could get good practice by proving this by induction. –  Jonas Meyer Jun 27 '12 at 20:48
    
I think sometimes it is better to explain where that comes from. It is of no use if the OP has no clue on how to deduce such inequality. –  Pedro Tamaroff Jun 28 '12 at 5:06
    
@Peter: How do we know whether or not the OP has a clue? No such information was given. –  Jonas Meyer Jun 28 '12 at 15:27
    
@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". –  Pedro Tamaroff Jun 28 '12 at 16:31

Here's another (somewhat roundabout) way to approach this. Noting that we've got $n!$ terms showing up, it suggest 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 $$\lim_{n\to\infty}\frac{2^n}{n!}=0\Longrightarrow \lim_{n\to\infty}\frac{n!}{2^n}=\infty$$

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