Prove $$\lim_{n \to \infty}\frac{a^n}{n!}=0~~~(a>0).$$
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.
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
$$n\log a-\sum_{k=1}^{n} \log k\sim -n\log n+n(1+\log a)\to -\infty$$
Notice that
$$\sum_{n=0}^\infty \frac{a^n}{n!} = \exp a \in \mathbb{R}$$
so the series in particular converges and $\frac{a^n}{n!} \to 0$.
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!
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.
