Understanding the solution to $\lim_{n\to\infty}\frac{a^n}{n!}, a>1$ Calculate $$\lim_{n\to\infty}\frac{a^n}{n!}, a>1$$
I need help understanding the solution to this problem:

$$a_n:=\frac{a^n}{n!}$$
$$a_{n+1}=\frac{a^{n+1}}{(n+1)!}=\frac{a}{n+1}\cdot \frac{a^n}{n!}=\frac{a}{n+1}\cdot a_n$$
For $$n\geq n_0 :=\left \lfloor{a}\right \rfloor +1>a$$ we have $$\frac{a}{n+1}<\frac{a}{a+1}<1$$
$\Rightarrow$ the sequence $a_n$ is decreasing from $n_0$-th term onwards and obviously $a_n\geq 0, \forall n\in \Bbb N \Rightarrow a_n$ is convergent.
Let $L:=\lim_{n\to\infty}a_n$. Then
$$a_{n+1}=\frac{a}{n+1}\cdot a_n$$
$$L=\lim_{n\to\infty}\frac{a}{n+1}$$
$$L=0\cdot L=0 \Rightarrow L=0$$

Why did they define $n_0 =\left \lfloor{a}\right \rfloor +1$? Furthermore, I don't understand what happened after they defined $L=\lim_{n\to\infty}a_n$. How did they get $\lim_{n\to\infty}\frac{a}{n+1}$ from $\frac{a}{n+1}\cdot a_n$?
 A: Using the idea of the comments, you can construct the proof in the following way: First choose $K = [a] + 1$, then rewrite: $0 <a_n = \dfrac{a}{1}\cdot \dfrac{a}{2}\cdots \dfrac{a}{K-1}\cdot\dfrac{a}{K}\dfrac{a}{K+1}\cdot \dfrac{a}{K+2}\cdots \dfrac{a}{n-1}\cdot \dfrac{a}{n}= C\cdot \dfrac{a}{K}\cdot \dfrac{a}{K+1}\cdots \dfrac{a}{n}\le C\left(\dfrac{a}{K}\right)^{n-K+1}$, and apply squeeze theorem to conclude since $0 < \dfrac{a}{K} < 1$.
A: If $L:=\lim_{n\to \infty} a_n$ exists then applying limits both sides we get
$$\lim a_{n+1}=\lim \frac{a}{n+1} a_n\implies L=L \cdot\lim\frac{a}{n+1}=L\cdot 0=0$$
The limit $L$ exists because $a_n$ is decreasing and bounded, I dont know if you know this theorem, named as monotone convergence theorem related to sequences of real numbers.
You can see that $\lim\frac{a}{n+1}=0$?

The thing related to $n_0=\lfloor a\rfloor +1$ is to show that $L$ exists because the sequence for $n\ge n_0$ is monotone (decreasing) and bounded (because $a_n>0$ for all $n\in\Bbb N$).
A: i think this term is missing: $$L=\lim_{n\to \infty}\frac{a}{n+1}\lim_{n \to \infty}a_n$$
