Using $\Gamma(z) = \lim_{n \to \infty} \left[ \frac{n! n^z}{z(z+1)(z+2)\ldots(z+n)} \right]$ to prove the Weiestrass product.

Question

I was searching the web quite thoroughly in the last two days. I was in paralytically looking for a rigorous proof using $$ \Gamma(z) = \lim_{n \to \infty} \left[ \frac{n! n^z}{z(z+1)(z+2)\ldots(z+n)} \right] $$
to prove the canonical wierstrass representation of the gamma function $$ \frac{1}{\Gamma(z)} = z e^{\gamma z} \prod_{n=1}^\infty \left( 1 + \frac{z}{n}\right) e^{-z/n} $$
I have seen a proof for the reverse, example here http://www.proofwiki.org/wiki/Equivalence_of_Gamma_Function_Definitions but how does one go about proving it the other way?

Answer 1

According to the limit, $$ \begin{align} \frac1{\Gamma(z)} &=z\lim_{n\to\infty}n^{-z}\prod_{k=1}^n\left(1+\frac zk\right)\\ &=z\lim_{n\to\infty}n^{-z}\prod_{k=1}^ne^{z/k}\prod_{k=1}^n\left(1+\frac zk\right)e^{-z/k}\\ &=z\lim_{n\to\infty}e^{z(H_n-\log(n))}\prod_{k=1}^n\left(1+\frac zk\right)e^{-z/k}\\ &=ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac zk\right)e^{-z/k}\\ \end{align} $$

Answer 2

Hint Write $${\Gamma _n}(z) = {z^{ - 1}}{e^{z\log n}}\prod\limits_{k = 1}^n {{{\left( {1 + \frac{z}{k}} \right)}^{ - 1}}} $$

and use that $$\gamma=-\lim\limits_{n\to\infty}\left(\log n-H_n\right)$$ to get$${\Gamma _n}(z) = {z^{ - 1}}{e^{-z\left( {{H_n} - \log n} \right)}}\prod\limits_{k = 1}^n {{{\left( {1 + \frac{z}{k}} \right)}^{ - 1}}{e^{ \frac{z}{k}}}} $$

Then let $n\to \infty$.