Prove $\lim\limits_{n \to \infty}\frac{a^n}{n!}=0~~~(a>0).$ Problem
Prove  $$\lim_{n \to \infty}\frac{a^n}{n!}=0~~~(a>0).$$
Proof
Denote $x_n=\dfrac{a^n}{n!}$ where $n=1,2,\cdots.$Then $$\frac{x_{n+1}}{x_n}=\frac{a^{n+1}}{(n+1)!}\cdot \frac{n!}{a^n}=\frac{a}{n+1}\to 0~~~(n \to \infty)$$which implies the series $\sum\limits_{n=1}^{\infty} x_n$ is convergent, by the ratio test. It follows that $$\lim_{n \to \infty}x_n=0,$$which is just the necessary condition for the positive convergent series.
 A: Yes it is correct, note also that the ratio test works for positive sequences and we don't need to consider the series, that is 

$$\lim_{n\to \infty}\frac{x_{n+1}}{x_n}=\begin{cases}L<1 \implies x_n\to 0\\L>1 \implies x_n\to \infty\\L=1 \quad \text{we can't conclude}\end{cases}$$


As an alternative, just to play around, we can use that
$$\frac{a^n}{n!}=e^{n\log a-\sum_{k=1}^{n} \log k} \to 0$$
indeed by integral approximation 
$$\sum_{k=1}^{n} \log k \sim \int_1^n \log x dx  =n\log n-n+1$$
and therefore
$$n\log a-\sum_{k=1}^{n} \log k\sim -n\log n+n(1+\log a)\to -\infty$$
A: Notice that 
$$\sum_{n=0}^\infty \frac{a^n}{n!} = \exp a \in \mathbb{R}$$
so the series in particular converges and $\frac{a^n}{n!} \to 0$.
A: Yes，we may give another proof by the squeeze theorem.
If $0<a<1.$ Then
$$0<\frac{a^n}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdots\frac{a}{n}<\frac{a}{n}\to 0~~~(n \to \infty);\tag1$$
If $a \geq 1.$ Then there exists an integer $k \geq 1$ such that $k \leq a < k+1.$ Thus 
$$0<\frac{a^n}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdots\frac{a}{k}\cdot\frac{a}{k+1}\cdots\frac{a}{n}<\frac{a^k}{k!}\cdot\frac{a}{n}\to 0~~~(n \to \infty).\tag2$$
By $(1)$ and $(2)$, we are done!
A: we can force a to lie between (k, k+1) and apply sandwich theorem.
A: Besides, we may also use the monotonic bounded principle.
Notice that $$x_{n+1}=x_n\cdot \frac{a}{n+1},\tag 1$$which implies that $\{x_n\}$ is decreasing when $n>a-1$. And $\{x_n\}$ has a lower bound, for $x_n>0$. Thus, $\{x_n\}$ has a limit, which can be denoted as $x$. Therefore, taking the limits of both sides of $(1)$, we obtain an equation $$x=x\cdot 0,$$which gives $x=0$. This is the limit what we want.
