Expressing Gamma function using Zeta series

Question

It is known that: $$ \zeta(1-x) = \sum_{n=1}^{\infty} \frac{1}{n^{1-x}} = \sum_{n=1}^{\infty} \frac{n^{x}}{n} \quad \text{for $x<0$} $$ Is it true that: $$ \Gamma(x) = \left( \sum_{n=1}^{\infty} \frac{n^{x}}{n} \right) \div \left( {\prod_{n=1}^{\infty} {(1+\frac{x}{n}})} \right) \quad \text{for $x>0$} $$ How to proof it?

Answer 1

using $H_N = \sum_{n=1}^N \frac{1}{n}$ and $ H_N-\gamma=\ln N+\mathcal{O}(1/N)$ and $\Gamma(z) = \frac{1}{z} e^{-\gamma z}\prod_{n=1}^\infty (1+\frac{z}{n})^{-1} e^{z/n}$ :

$$\Gamma(z) = \lim_{N \to \infty} \frac{1}{z} e^{-\gamma z}\prod_{n=1}^N (1+\frac{z}{n})^{-1} e^{z/n} =\lim_{N \to \infty} \frac{1}{z} e^{(H_N-\gamma) z}\prod_{n=1}^N (1+\frac{z}{n})^{-1}$$ $$=\lim_{N \to \infty} \frac{1}{z} e^{(\ln N+\mathcal{O}(1/N)) z}\prod_{n=1}^N (1+\frac{z}{n})^{-1} = \lim_{N \to \infty} \frac{1}{z} N^z\prod_{n=1}^N (1+\frac{z}{n})^{-1}$$

and for $Re(z) > 0$, $\ \sum_{n=1}^N n^{z-1} = \int_1^N x^{z-1}dx + \mathcal{O}(N^{z-1}) =\frac{N^z}{z}+\mathcal{O}(N^{z-1})$ : $$\boxed{\Gamma(z) = \lim_{N \to \infty} \sum_{n=1}^N n^{z-1}\prod_{n=1}^N (1+\frac{z}{n})^{-1} \qquad Re(z) > 0\ }$$