# How to prove $\lim_{n\rightarrow \infty} {a^n \over n!}=0$ [duplicate]

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

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

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

• A similar question: this one. Commented Mar 17, 2012 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. Commented Mar 17, 2012 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
Commented Mar 17, 2012 at 17:42
• @Michael, I agree, but rather than suggesting that all these answerers upvote, I'd say that they shouldn't have answered... Commented Mar 17, 2012 at 22:23
• @robjohn I'd say it's more than possible! Commented Mar 18, 2012 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 $\frac{a^n}{n!} \to 0$.

• 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. Commented Mar 17, 2012 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.

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 $0$.

• Nice quoting of a principle that goes back to Archimedes. Commented Mar 17, 2012 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. Commented Mar 17, 2012 at 22:53
• I was complimenting, and upvoting. Commented Mar 18, 2012 at 0:49

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

• That certainly works, but it's a far more sophisticated tool than what is needed. Commented Mar 17, 2012 at 16:54
• If someone asked this question, he shouldn't know this result... Commented Mar 17, 2012 at 16:59
• Yet, this is basically "the reason" the limit is zero. The numerator grows exponentially, but the denominator is growing like $c^{n \log n}$. Commented Mar 17, 2012 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.

$n! > (n/2)^{n/2}$ 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 $n > 4a^2$.

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