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

As the topics, how to prove $\lim\limits_{n\rightarrow \infty} \dfrac{a^n}{n!}=0$

$\forall a \in \mathbb{R^+}$

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    @robjohn: I saw your comment.2012-03-18

7 Answers 7

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I don't know if you're allowed to use this but you could argue as follows:

$ \lim_{N \to \infty} \sum_{n=0}^N \frac{a^n}{n!} = e^a < \infty$

hence $\frac{a^n}{n!} \to 0$.

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    You're certainly "allowed to use it" unless there's some particular reason why you're not (e.g. your audience doesn't know it, or it's homework and that constraint was handed down from above). But the question has an easy answer that doesn't require knowledge of this series.2012-03-17
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If $n>2|a|$, then every time you increment $n$ by $1$, you're making the value of the fraction less than half what it was. If you cut something down to less than half its previous size at each step, then its size approaches $0$.

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    I was complimenting, and upvoting.2012-03-18
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Let's suppose $a$ is... I don't know, $5$ for a moment. And let's look at the sequence.

$\dfrac{5}{1}, \dfrac{5^2}{2\cdot 1}, \dfrac{5^3}{3\cdot 2}, \dfrac{5^4}{4 \cdot 3 \cdot 2}, \dfrac{(5) \cdot 5^4}{(5) \cdot 4 \cdot 3 \cdot 2}, \dfrac{(5^2) \cdot 5^4}{(6 \cdot 5) \cdot 4!}, \dfrac{(5^3) \cdot 5^4}{(7 \cdot 6 \cdot 5) 4!}, \ldots$

So in particular, after $n = 5$, we have a constant multiplied by something bounded by $\left(\frac{5}{6}\right)^n$, and thus it goes to 0.

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Hint: for large $n$, use stirling approximation for factorial.

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    Yet, this is basically "the reason" the limit is zero. The numerator grows exponentially, but the denominator is growing like $c^{n \log n}$.2012-03-17
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Hint : You can show that : $c_n = \frac{a^n}{n!}$ and compute : $\frac{c_{n-1}}{c_n} = \frac{n}{a} \rightarrow \infty$ and conclude.

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$n! > (n/2)^{n/2}$ so $a^n/n! < a^n/(n/2)^{n/2} = (a^2)^{n/2}/(n/2)^{n/2} = (2a^2/n)^{n/2} < (1/2)^{n/2}$ for $n > 4a^2$.

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Let $k$ be an integer so that $k>a$. You can take for example $k= \lfloor a \rfloor +1$.

Let $C=\frac{a^k}{k!}$, which is a constant.

Claim: for $n \geq k+1$ we have

$0 \leq \frac{a^n}{n!} \leq \frac{aC}{n} $

The left inequality is clear, while the RHS is

$\frac{a^n}{n!} = \frac{a^k}{k!}\frac{a}{k+1}\frac{a}{k+1}...\frac{a}{n} \leq C \cdot 1 \cdot 1 ... \cdot 1 \cdot \frac{a}{n} \,.$

Now Squeeze and you are done. Or if you know how to deal with $\epsilon$, pick an $N_\epsilon > \frac{aC}{\epsilon}$