Calculate $\lim_{n \to \infty}\frac{a^n}{n!}$ Calculate $$\lim_{n \to \infty}\frac{a^n}{n!}$$
Attempt
Consider $$\lim_{n \to \infty}\exp(\log(\frac{a^n}{n!}))=\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)$$
for $a>1$
$$ \lim_{n \to \infty}\left(n\log(a)\right)=\infty $$
$$ \sum_{n \geq 1}\log(n)\leq \int_{1}^{\infty}{\log(x)dx}=\infty$$
therefore $$\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)=\exp(-\infty)=0$$
therefore
$$\lim_{n \to \infty}\frac{a^n}{n!}=0$$
Is my proof right?
I think that is not general, since $a>1$ and in the problem $a$ is arbitrary
 A: Replying to your last comment: Yes there is a faster way. Show that for sufficiently large $n$ the following inequality holds:
$$n!>(a+1)^n.$$
We now have
$$\lim_{n\to\infty} \frac{a^n}{n!} \leq \lim_{n\to\infty} \frac{a^n}{(a+1)^n} = \lim_{n\to\infty} \left(\frac{a}{a+1}\right)^n \to 0.$$
A: Here is an alternative:
It seems you are working with $a>1$, although it is not stated at the beginning but rather in the middle. I will go with that assumption:
$a^n>0$ and $n! > 0, \forall n \in \mathbb{N}$.
Now,

$\displaystyle \frac{a^n}{n!} = \frac{a\cdot a \cdots a}{1\cdot 2 \cdots n} $

At some point we will have $n > a$, say this occurs at $N$, take $n > N$ then

$\displaystyle \frac{a^n}{n!} = \frac{a\cdot a \cdots a}{1\cdot 2 \cdots n}  \le \frac{a \cdots a}{1 \cdots (N-1)}\cdot \frac{a \cdots a}{N \cdots n} \le  \frac{a^{N-1}}{(N-1)!}\cdot \frac{a^{n-(N-1)}}{N\cdots n}$

Now, since we have $m > a, \forall m \in \{N,\ldots,n\}$ we know $\displaystyle \frac{a}{m} < 1$
Hence

$\displaystyle \frac{a^{N-1}}{(N-1)!}\cdot \frac{a^{n-(N-1)}}{N\cdots n} \le \frac{a^{N-1}}{(N-1)!}\cdot \frac{a}{n} \to 0$ as $n \to \infty$

A: Hint : Starting from your idea, try to find the limit of $n log(a) - \sum_{k=1}^n log(k)$. Then apply what you know about the $exp$ function.
