Tag Archives: product

Weierstrass representation of Gamma function proof

$$\Gamma(z)\,= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}$$ $\textit{where } \gamma \textit{ is the Euler constant}$ $$\textit{proof}$$ Take logarithm to the Euler representation $$\log z\Gamma(z) =\lim_{n\to \infty}z\sum_{k=1}^n \left(\log\left(1+k\right)-\log(k)\right)-\sum_{k=1}^n\log\left(1+\frac{z}{k}\right)$$ Note the alternating sum $$\sum_{k=1}^n \left(\log\left(1+k\right)-\log(k)\right) = \log(n+1)$$ Hence we have … Continue reading

$$\Gamma(z) = \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = \frac{1}{z}\prod^\infty_{k=1}\frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}}$$ $$\textit{proof}$$ Note that $$\Gamma(z+n+1) = \Gamma(z+1) \prod^{n}_{k=1}(k+z)$$ Which indicates that $$\prod^{n}_{k=1}(k+z) = \frac{\Gamma(z+n+1)}{z\Gamma(z)}$$ Also note that $$\prod^{n}_{k=1}k = n!$$ Hence we have \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = … Continue reading