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What sequence is dominant between

$ f(n) = n!$ and
$ g(n) = 2^n$ or (or $a^n$)

I mean $ f/g -> 0 $ or $infinity$

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2 Answers 2

up vote 1 down vote accepted

Stirling's approximation states that:

$$ \lim_{n \rightarrow \infty} {\frac{n!}{\sqrt{2\pi n}\, \left(\frac{n}{e}\right)^n}} = 1 $$


$$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n $$

Which means that $n!$ grows faster that $a^n$. This is easy to see intuitively, given how $a^n$ consists of $n$ constant terms, whereas $n!$ consists of $n$ increasing terms. For large enough $n$, $n!$ will be larger.

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The factorial grows faster because it consists of $n$ factors of increasing size, while the power $a^n$ consists of $n$ factors of constant size.

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you could give a simple proof, something like n! > (n/4)^n for n>5 maybe – kwak May 30 '12 at 13:52
well I see your point n! is > a^n for n>a/2 right? – kwak May 30 '12 at 13:53

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