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

$$\log z\Gamma(z) =\lim_{n\to \infty}z\log(n+1)-\sum_{k=1}^n\log\left(1+\frac{z}{k}\right)$$

Now we can use the harmonic numbers

$$H_n = \sum_{k=1}^n \frac{1}{n}$$

Add and subtract $zH_{n+1}$

$$\log z\Gamma(z) =\lim_{n\to \infty}z\log(n+1)-zH_{n+1}+\sum_{k=1}^n\left[\log\left(1+\frac{z}{k}\right)^{-1}+\frac{z}{k}\right] +\frac{z}{n+1}$$

The last term goes to zero and by definition we have the Euler constant

$$\gamma = \lim_{n \to \infty }H_n -\log(n)$$

Hence the first term is the Euler constant

$$\log z\Gamma(z) =-z\gamma+\sum_{k=1}^\infty \log\left(1+\frac{z}{k}\right)^{-1}+\frac{z}{k}$$

By taking the exponent of both sides

$$z\Gamma(z) =e^{-\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)^{-1}e^{\frac{z}{k}}$$

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