prove-that-1-z-z-e-z-n-1-1-z-n-e-z-n-

Question Number 67537 by mathmax by abdo last updated on 28/Aug/19

p r o v e t h a t \frac{1}{Γ (z)} = z e^{γ z} \prod_{n = 1}^{\infty} (1 + \frac{z}{n}) e^{- \frac{z}{n}}

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

L e t u s e t h e r e s u l t \frac{1}{Γ (z)} = \lim_{n \to \infty} \frac{z (z + 1) \dots (z + n)}{n^{z} n!}

\frac{1}{Γ (z)} = \lim_{n \to \infty} {z e}^{- z l n (n)} \frac{\underset{k = 1}{\prod^{n}} (z + k)}{\underset{k = 1}{\prod^{n}} k} = \lim_{n \to \infty} {z e}^{- z l n (n)} \underset{k = 1}{\prod^{n}} [(1 + \frac{z}{k}) e^{\frac{- z}{k}} . e^{\frac{z}{k}}]

= \lim_{n \to \infty} z \underset{k = 1}{\prod^{n}} [(1 + \frac{z}{k}) e^{- \frac{z}{k}}] e^{z (- l n (n) + \underset{k = 1}{\sum^{n}} \frac{1}{k})}

= z \underset{k = 1}{\prod^{\infty}} [(1 + \frac{z}{k}) e^{- \frac{z}{k}}] e^{z γ} c a u s e \lim_{n \to \infty} \underset{k = 1}{\sum^{\infty}} \frac{1}{k} - l n (n) = γ