Deduce that $\Gamma(x)=\lim_{n\to\infty}\frac{n!x^n}{x(x+1)\cdots(x+n)}$ for all $x$, given that it holds for $0<x<1$

Question

How to understand rigorously that $(95)$ hold for any $x>0$ using that $\Gamma(x+1)=x\Gamma(x)$. Can anyone explain please this strictly?

Answer 1

For fixed $y\in(0,1]$, by induction on $m\geq 0$. Prove:

$$\Gamma(m+y)=\lim_{n\to \infty} \frac{n!n^{m+y}}{(m+y)(m+y+1)(m+1+2)\dots(m+y+n)}$$

You already know if for $m=0$ if $y\neq 1$. Prove it for $y=1, m=0$, too.

Then assume true for $m$.

Then $$\begin{align}\Gamma(m+1+y)=(m+y)\Gamma(m+y)&=(m+y)\lim_{n\to\infty}\frac{n!n^{m+y}}{(m+y)\cdots(m+y+n)}\\ &=\lim_{n\to\infty}\frac{n!n^{m+y}}{(m+y+1)(m+y+2)\cdots(m+y+n)} \end{align}$$

Letting $k=n-1$, we get:

$$\Gamma(m+1+y)=\lim_{k\to\infty} \frac{(k+1)!(k+1)^{m+y}}{(m+y+1)(m+y+2)\cdots(m+1+y+k)}$$

But $$(k+1)!(k+1)^{m+y} = k!(k+1)^{m+1+y},$$ and:

$$\lim_{k\to\infty} \frac{(k+1)^{m+1+y}}{k^{m+1+y}}=1$$

So we get:

$$\Gamma(m+1+y)=\lim_{k\to\infty} \frac{k!k^{m+y+1}}{(m+y+1)(m+y+2)\cdots(m+1+y+k)}$$

Which is what we were trying to prove.

Now, every $x>0$ can be written $x=y+m$ for some $y\in(0,1]$ and $m\geq 0$ an integer.

Basically, this is proving by induction on $m=\lceil x\rceil-1$.

Answer 2

We know that $\Gamma(x)=\int_{0}^{+\infty}e^{-t}t^{x-1}dt$. Also $$ e^{-t}=\lim_{n\rightarrow +\infty }\Big(1-\frac{t}{n}\Big)^{n}. $$ Hence we can write $\Gamma(x)=\lim_{n\rightarrow +\infty}\Gamma_{n}(x)$ where $$ \Gamma_{n}(x)=\int_{0}^{n}\big(1-t/n\big)^{n}t^{x-1}dt= n^{x}\int_{0}^{1}(1-u)^{n}u^{x-1}du, \quad n\geq 1. $$ where the last equality obtained by using $t=nu$. Then you can easily see that $\Gamma_{1}(x)=1/(x(x+1))$ , $\Gamma_{2}(x)=2(2^{x})/(x(x+1)(x+2))$ and generally, by $n$ times integration you will arrive at $$ \Gamma_{n}(x)=\frac{n^{x}n!}{x(x+1)...(x+n)}. $$