# Prove that for $a > 0$, $\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}$ [duplicate]

Possible Duplicate:
How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$

Prove that for $a > 0$, $\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}$. My attempt is since $$e^x=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$ and the series converges for all $x \in \mathbb{R}$, by the test of divergence, $$\lim_{n \rightarrow \infty}{\frac{a^n}{n!}=0}.$$ Is my proof correct ? or are there any alternative ?

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## marked as duplicate by David Mitra, Jonas Meyer, dtldarek, Henry T. Horton, MicahFeb 1 '13 at 16:11

Yes, it is correct. – Siminore Feb 1 '13 at 15:40
It's correct, but uses a "sledgehammer" in my opinion. See this post and its links for more elementary methods. – David Mitra Feb 1 '13 at 15:41
This is also true for $a\leq 0$. – 1015 Feb 1 '13 at 15:42
The proof is correct but, as @DavidMitra says, you're using a sledgehammer :) So I don't find it very clean, you should be able to prove then that those series are equal to $e^x$ and that they converge, using taylor's theorem also does the trick but it's more advanced that the analysis corresponding to this. Final message: I would look for another proof, but if I saw this in an exam and nothing came to my mind, I would undoubtedly use yours and it would be correct. – MyUserIsThis Feb 1 '13 at 15:49
@DavidMitra : What do you mean ' sledgehammer' in this case ? – Idonknow Feb 1 '13 at 15:53

$$\frac{a^n}{n!} = \frac{a}{1}\frac{a}{2}\frac{a}{3}\ldots\frac{a}{n}$$
The exponential is simply the repeated product of $n$ number of $a$'s, whereas the factorial is the repeated product of $n$ ever-growing integers.
Therefore there must come a point, some $n \ge a$, after which all new factors of the denominator must be larger than the new factors of the numerator. This gap in the size of $a$ and $n$ will only continue to grow; the final factor itself, $\frac{a}{n}$, goes to $0$ as $n$ does, and so the product of an infinite amount of such small fractions has little choice but to follow suit and approach $0$.