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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?

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    $\begingroup$ $\prod_{n \ge 1} 1+ \frac{z}{n}$ 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}$) $\endgroup$ – reuns Sep 7 '16 at 22:00
  • $\begingroup$ The expression is a limit of indefinite case ($\infty \div \infty$). So, is the limit do exist? $\endgroup$ – Hazem Orabi Sep 7 '16 at 22:11
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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 $Re(z) > 0$, $\ \sum_{n=1}^N n^{z-1} = \int_1^N x^{z-1}dx + \mathcal{O}(N^{z-1}) =\frac{N^z}{z}+\mathcal{O}(N^{z-1})$ : $$\boxed{\Gamma(z) = \lim_{N \to \infty} \sum_{n=1}^N n^{z-1}\prod_{n=1}^N (1+\frac{z}{n})^{-1} \qquad Re(z) > 0\ }$$

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  • $\begingroup$ If the expressing is valid for x>0, Is it okay to be re-written as: $$ \Gamma(x) = \prod_{p \space prime} \frac{1}{1-1/{p^{1-x}}} \space \prod_{n=1}^{\infty} {\left(1+\frac{x}{n}\right)}^{-1} \quad \text{for $0<x<1$} $$ $\endgroup$ – Hazem Orabi Sep 7 '16 at 23:18
  • $\begingroup$ @HazemOrabi no, read an introduction to analysis course $\endgroup$ – reuns Sep 8 '16 at 14:58
  • $\begingroup$ 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: $$\frac{N^z}{z} \rightarrow \sum_{n=1}^N {n^{z-1}} + \mathcal{O}(N^{z-1})$$ $\endgroup$ – Hazem Orabi Sep 8 '16 at 15:23
  • $\begingroup$ @HazemOrabi read an analysis course $\endgroup$ – reuns Sep 8 '16 at 15:24

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