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I'm having some trouble proving that this sequence diverges. Can anyone suggest how to approach & proceed?

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

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    $\begingroup$ consider $a_n = \frac12 \frac22 \frac 32 \frac 42 \ldots \frac{n}2$; or $a_{2n} = \frac12 \frac22 \frac 32 \frac 42 \ldots \frac{2n}2$; or the fact that $a_{n+1} = \frac{n+1}{2} a_n$ $\endgroup$
    – user66081
    Jan 16, 2014 at 4:09

3 Answers 3

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Note that $n! \geq n\cdot 2^{n-2}$ (at least for large $n$)

Thus $$ a_n = \frac{n!}{2^n} \geq \frac{n2^{n-2}}{2^n} = n\cdot\frac{1}{4} \stackrel{n\to\infty}{\longrightarrow}\infty$$

Which means $a_n$ tends to $\infty$

Where $n! \geq n\cdot 2^{n-2}$ holds because

$$n! = n\cdot\underbrace{\underbrace{(n-1)}_{\geq2}\cdot\underbrace{(n-2)}_{\geq2}\dots\cdot\underbrace{3}_{\geq2}\cdot\underbrace{2}_{\geq2}}_{n-2 \text{ factors }}\cdot1$$

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  • $\begingroup$ Thank you! I understand, but where does the first line come from? $\endgroup$
    – user120494
    Jan 16, 2014 at 4:18
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    $\begingroup$ I made an edit so you can see it $\endgroup$
    – user127.0.0.1
    Jan 16, 2014 at 4:22
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Note that $a_n \geq 1$ for all n. Also, $a_{n+1}/a_n=(n+1)/2$. So it's clear that $a_n$ can't converge, as otherwise $a_{n+1}/a_n$ would have limit $1$ while $(n+1)/2$ diverges.

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$n!/2^n=(\frac{1}{2})(\frac{2}{2})(\frac{3}{2})\cdots (\frac{n}{2})$. Except for the first two factors, these are all at least $\frac{3}{2}$, so the product is at least $(\frac{1}{2})(\frac{2}{2})\cdot (\frac{3}{2})^{n-2}$ for $n>2$.

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