Show that the gamma function converges

Question

One way to define the gamma function is $$\Gamma(x)= \lim_{p\rightarrow \infty} \Gamma_p(x) $$

Where $$\Gamma_p(x)=\dfrac{p! p^x }{x(x+1)...(x+p)}=\dfrac{p^x}{x(1+x/1)(1+x/2)...(1+x/p)} $$

How to prove that the limit converges?

Answer 1

Write this as

$$\Gamma_p(x) = \left[xe^{-x \ln p}\prod_{k=1}^p\left(1+\frac{x}{k}\right)\right]^{-1}=\left[xe^{x (\sum_{k=1}^p1/k-\ln p)}\prod_{k=1}^p\left(1+\frac{x}{k}\right)e^{-x/k}\right]^{-1}.$$

We show the limit as $p \to \infty$ exists by examining the convergence of the sum and product appearing in this expression..

Note that

$$\lim_{p \to \infty} \left(\sum_{k=1}^{p}\frac{1}{k} - \ln p \right)= \gamma,$$

where $\gamma$ is the Euler-Mascheroni constant.

The product

$$\prod_{k=1}^{p} \left(1 + \frac{1}{k}\right)e^{-x/k}$$

converges if and only if the following series converges

$$\sum_{k=1}^{\infty} \left[\ln\left(1+\frac{x}{k}\right) - \frac{x}{k} \right].$$

The series does indeed converge as the summand is asymptotic to $x^2/(2k^2)$ as $k \to \infty$.

Hence, the original sequence converges to the Weierstrass form of the gamma function:

$$\Gamma(x) =\left[xe^{\gamma x }\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-x/k}\right]^{-1}$$