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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 $\frac{a}{n+1}\cdot a_n$?

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    $L = \lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac a{n+1}$ (which goes to zero, because $a$ is fixed, and $n+1$ keeps increasing), so that's how they get zero. The first $n_0$ is chosen in a manner such that after the $n_0$th term, the sequence is decreasing, because after that, the ratio of successive terms is smaller than $1$. – 2017-01-07
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    Perhaps there was a typo in the original? In the line just before the bottom line, it should read $L = \lim_{n \rightarrow \infty} \frac{a}{n+1} L$ (instead of $L = \lim_{n\rightarrow \infty} \frac{a}{n+1}$) which explains the bottom line. – 2017-01-07

3 Answers 3

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i think this term is missing: $$L=\lim_{n\to \infty}\frac{a}{n+1}\lim_{n \to \infty}a_n$$

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Using the idea of the comments, you can construct the proof in the following way: First choose $K = [a] + 1$, then rewrite: $0

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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 $n\in\Bbb N$).