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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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marked as duplicate by robjohn, Rudy the Reindeer, Dylan Moreland, Américo Tavares, t.b. Mar 18 '12 at 17:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

A similar question: this one. – Américo Tavares Mar 17 '12 at 16:56
Five answers have been posted, and I'm still the only person who's up-voted this question. I tend to think that if a question is worth answering, then it's worth voting for. – Michael Hardy Mar 17 '12 at 17:04
@Michael: I disagree. A question could be very interesting, even if the one asking the question did not even try to answer the question. If anything, I would downvote this question, since "The question does not show any research effort". – TMM Mar 17 '12 at 17:42
@Michael, I agree, but rather than suggesting that all these answerers upvote, I'd say that they shouldn't have answered... – The Chaz 2.0 Mar 17 '12 at 22:23
@robjohn I'd say it's more than possible! – Dylan Moreland Mar 18 '12 at 17:35

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. – Michael Hardy Mar 17 '12 at 17:02

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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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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Nice quoting of a principle that goes back to Archimedes. – André Nicolas Mar 17 '12 at 17:38
@AndréNicolas : I'm not sure if you're complimenting me on this answer or complaining that it's plagiarized from Archimedes, or what. But if I wanted to explain what Archimedes said, I'd certainly be a lot more explicit and phrase it differently. However, this doesn't seem like the right context for that. – Michael Hardy Mar 17 '12 at 22:53
I was complimenting, and upvoting. – André Nicolas Mar 18 '12 at 0:49

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

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That certainly works, but it's a far more sophisticated tool than what is needed. – Michael Hardy Mar 17 '12 at 16:54
If someone asked this question, he shouldn't know this result... – Adrien Boulanger Mar 17 '12 at 16:59
Yet, this is basically "the reason" the limit is zero. The numerator grows exponentially, but the denominator is growing like $c^{n \log n}$. – David Harris Mar 17 '12 at 19:51

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 $N_\epsilon > \frac{aC}{\epsilon}$

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