# Expressing Gamma function using Zeta series

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?

• $\prod_{n \ge 1} 1+ \frac{z}{n}$ doesn't converge (that's why $\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{n\ge 1} \left(1 + \frac{z}{n}\right)^{-1} e^{\frac{z}{n}}$ and not $\frac{1}{z}\prod_{n \ge 1} (1+ \frac{z}{n})^{-1}$) – reuns Sep 7 '16 at 22:00
• The expression is a limit of indefinite case ($\infty \div \infty$). So, is the limit do exist? – Hazem Orabi Sep 7 '16 at 22:11

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 $Re(z) > 0$, $\ \sum_{n=1}^N n^{z-1} = \int_1^N x^{z-1}dx + \mathcal{O}(N^{z-1}) =\frac{N^z}{z}+\mathcal{O}(N^{z-1})$ : $$\boxed{\Gamma(z) = \lim_{N \to \infty} \sum_{n=1}^N n^{z-1}\prod_{n=1}^N (1+\frac{z}{n})^{-1} \qquad Re(z) > 0\ }$$
• If the expressing is valid for x>0, Is it okay to be re-written as: $$\Gamma(x) = \prod_{p \space prime} \frac{1}{1-1/{p^{1-x}}} \space \prod_{n=1}^{\infty} {\left(1+\frac{x}{n}\right)}^{-1} \quad \text{for 0<x<1}$$ – Hazem Orabi Sep 7 '16 at 23:18
• Thank you. I accept your first answer. Nevertheless, (1) We may start immediately from Euler definition: $$\Gamma(z) = \frac{1}{z} \space \prod_{n=1}^{\infty} \frac{{\left(1+\frac{1}{n}\right)}^{z}} {1+\frac{z}{n}}$$ (2) We need a clear path to calculate the limit after the substitution: $$\frac{N^z}{z} \rightarrow \sum_{n=1}^N {n^{z-1}} + \mathcal{O}(N^{z-1})$$ – Hazem Orabi Sep 8 '16 at 15:23