Proving $\Gamma(x)$ is holomorphic

Question

My professor defined Gamma function in the following way:$$\Gamma(z)= \lim \limits_{n \rightarrow \infty} \frac{n!n^z}{z(z+1)....(z+n)}$$ Now we first observe that $f_n(z)= \frac{n!n^z}{z(z+1)....(z+n)}$ is holomorphic in ${\operatorname{Re} z>0}$. Then we have to show that the limit exists and also that $f_n$ converge uniformly. But I am unable to show that limit exists also I cannot prove that $f_n$ converge uniformly. If anyone can give a hint, it would be really great. Thank you.

Answer 1

Writing the limit for $\Gamma(z+1)$ (which is a tad easier) we get:

$$\lim_{n\to\infty}\left({n\over n+1}\right)^z\prod_{j=1}^n\left(1+{z\over j}\right)^{-1}\left(1+{1\over j}\right)^z\qquad (*)$$

we see that

$$\left(1+{z\over j}\right)^{-1}\left(1+{1\over j}\right)^z=1+{z(z-1)\over 2j^2}+O\left(j^{-3}\right)$$

implies convergence of the product and existence of the limit. (I assume you're familiar with infinite products, if not please say so and I'll edit). You can see uniform convergence in the desired region easily from this estimate as well.

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In case you cannot see it, here is a proof of $(*)$:

Note that

$$\left(1+{z\over j}\right)^{-1}=\left({j\over z+j}\right)$$

so that

$$\prod_{j=1}^n\left(1+{z\over j}\right)^{-1}={n!\over (z+1)(z+2)\ldots (z+n)}$$

and

$$\left(1+{1\over j}\right)^z=\left({j+1\over j}\right)^z$$

so that

$$\prod_{j=1}^n \left(1+{1\over j}\right)^z$$

is a telescoping product with value $(n+1)^z$.