Alternative definition of Gamma function. Show that $ \lim_{n \to \infty} \frac{n! \; n^m}{m \times (m+1) \times \dots \times (m+n)} = (m-1)!$

Question

Alternative definition of Gamma function on Wikipedia has it defined as a limit.

$$ \Gamma(t) = \lim_{n \to \infty} \frac{n! \; n^t}{t \times (t+1) \times \dots \times (t+n)}$$

How do we recover familiar properties of the Gamma function this way? Can we show that:

$$ \Gamma(3) = \lim_{n \to \infty} \frac{n! \; n^3}{3 \times 4 \times \dots \times (3+n)} \stackrel{?}{=} 2$$

I would like to see all integers. In that case we are showing that $\Gamma(m+1) = m!$

$$ \Gamma(m) = \lim_{n \to \infty} \frac{n! \; n^m}{m \times (m+1) \times \dots \times (m+n)} \stackrel{?}{=} 1 \times 2 \times 3 \times \dots \times (m-1)$$

What's throwing me off is that the factorial already appears on the bottom and somehow it shifts to the top.

Answer 1

Notice that:

$\Gamma(m) = \lim\limits_{n \to \infty} \frac{n! \; n^m}{m \times (m+1) \times \dots \times (m+n)} = \lim\limits_{n \to \infty} \frac{(m-1)! \; n! \; n^m}{(m + n)!} = (m - 1)! \times \lim\limits_{n \to \infty} \frac{n! \; n^m}{(m + n)!}$.

Now let's show that $\lim\limits_{n \to \infty} \frac{n! \; n^m}{(m + n)!} = 1$:

$\lim\limits_{n \to \infty} \frac{n! \; n^m}{(m + n)!} = \lim\limits_{n \to \infty} \frac{n^m}{(n + 1) \times \ldots \times (n + m)} = \lim\limits_{n \to \infty} \frac{1}{(1 + \frac{1}{n}) \times \ldots \times (1 + \frac{m}{n})} = 1$.

So,

$\Gamma(m) = (m - 1)! \times \lim\limits_{n \to \infty} \frac{n! \; n^m}{(m + n)!} = (m - 1)!$, q.e.d.

Answer 2

It can be proved by mathematical induction, it is easy to show that for $m=1$, $\Gamma (1)=1$, Suppose $m=k+1$, then $$\Gamma(k+1)=\lim _{n\rightarrow \infty}\frac{n!n^{k+1}}{\prod _{i=k+1}^{k+1+n}i}=\lim _{n\rightarrow \infty}\frac{n!n^{k}}{\prod _{i=k}^{k+n}i}\cdot \frac{nk}{k+1+n}$$, Let $$a_n=\frac{n!n^{k}}{\prod _{i=k}^{k+n}i}$$ and $$b_n=\frac{nk}{k+1+n}$$, therefore we have $\lim_{n\rightarrow \infty}a_n=\Gamma(k)=(k-1)!$ and $\lim_{n\rightarrow\infty}b_n=k$, so we have $\lim_{n\rightarrow \infty}a_nb_n=k!$

Answer 3

I just love little challenges...

$\Gamma(t)=\lim \limits_{n \to \infty}\frac{n!n^t}{t(t+1)(t+2)\ldots(t+n)}$

Note the denominator of the fraction is in itself a factorial of sorts.

$\lim \limits_{n \to \infty}\frac{n!n^t}{t(t+1)(t+2)\ldots(t+n)}=\lim \limits_{n \to \infty}\frac{n!n^t}{\frac{(t+n)!}{(t-1)!}}=\lim \limits_{n \to \infty}\frac{(t-1)!n!n^t}{(t+n)!}=\Gamma(t)=(t-1)!$

$\lim \limits_{n \to \infty}\frac{(t-1)!n!n^t}{(t+n)!}=(t-1)!$

$\lim \limits_{n \to \infty}\frac{n!n^t}{(t+n)!}=1$

$\lim \limits_{n \to \infty}n!n^t=\lim \limits_{n \to \infty}(t+n)!=\lim \limits_{n \to \infty}1 \times 2 \times 3 \times \ldots (n-1)(n)(n+1) \ldots (n+t-1)(n+t)$

$=\lim \limits_{n \to \infty}n!(n+1)(n+2)(n+3)\ldots(n+t-1)(n+t)$

$\lim \limits_{n \to \infty}n^t=\lim \limits_{n \to \infty}(n+1)(n+2)(n+3)\ldots(n+t-1)(n+t)$

$=\lim \limits_{n \to \infty}(t+n)(t+(n-1))(t+(n-2))\ldots(t+1)(t)=\lim \limits_{n \to \infty}(t)(t+1)(t+2)\ldots(t+(n-1))(t+n)$

From here, I have no idea. Whatever methods Euler or Weierstrass used to prove this must have been amazing.

I recall one of the famous things Euler did (The Basel Problem) was proving $\sum\limits_{n=1}^{\infty}\frac 1{n^2}=\frac{\pi^2}{6}$. It involved taylor series and multiplying a lot of terms out using Newton's Identities or something like that. I don't know if that is possible here, but I'm not $that$ good at math.