How to derive that $\lim_{n\to\infty}\frac{n!}{(n-a)!} = {1\over{(-a)}!}$

Question

According to Wolfram alpha:

$$\lim_{n\to\infty}\frac{n!}{(n-a)!} = {1\over{(-a)}!}$$

Is this correct? How can we arrive at this equality? I would think that this limit diverges since $ \frac{n!}{(n-a)!} = n(n-1)(n-2)...(n-a+1) $ which should diverge as $n$ goes to infinity?

Answer 1

I don't know how you got that answer, but Wolfram Alpha does not say that. It says

$$\lim_{n\to 0}\frac{n!}{(n-a)!} = \frac{1}{(-a)!}.$$

Rather, we have

$$\lim_{n \to \infty}\frac{n!}{(n-a)!} = \infty$$

if $a$ is positive, and

$$\lim_{n \to \infty}\frac{n!}{(n-a)!} = 0$$

if $a$ is negative.

If $a=0$ then the limit is $1$.