- Prove that
\[ \Gamma (s) = \lim _ {n \to \infty} \frac {n ^ {s} n !}{s (s + 1) \cdots (s + n)} \]
whenever \(s \neq 0 , - 1 , - 2 , . . .\)
[Hint: Use the product formula for \(1 / \Gamma\) , and the definition of the Euler constant γ.]
Solution
Goal: Prove Gauss’s limit formula for the Gamma function: \[ \Gamma(s) = \lim_{n \to \infty} \frac{n^s n!}{s(s+1)(s+2)\cdots(s+n)} \qquad \text{for } s \in \mathbb{C} \setminus \{0, -1, -2, \dots\}. \]
1. The Weierstrass product formula for the reciprocal Gamma function:Proof:
1.1. For any \(s \in \mathbb{C} \setminus \{0, -1, -2, \dots\}\), the reciprocal Gamma function is given by the entire product\[\frac{1}{\Gamma(s)} = s \, e^{\gamma s} \prod_{k=1}^\infty \left(1 + \frac{s}{k}\right) e^{-s/k},\]where \(\gamma = \lim_{n \to \infty} \left( H_n - \log n \right)\) is the Euler–Mascheroni constant and \(H_n = \sum_{k=1}^n \frac{1}{k}\).
1.2. Define the \(n\)-th partial product:\[P_n(s) = \prod_{k=1}^n \left(1 + \frac{s}{k}\right) e^{-s/k}.\]
1.3. By definition of the infinite product, \(\lim_{n \to \infty} P_n(s) = \frac{e^{-\gamma s}}{s \Gamma(s)}\).
2. Algebraic simplification of the partial product:Proof:
2.1. The product of linear factors simplifies to:\[\prod_{k=1}^n \left(1 + \frac{s}{k}\right) = \prod_{k=1}^n \frac{s + k}{k} = \frac{(s+1)(s+2)\cdots(s+n)}{n!} = \frac{s(s+1)(s+2)\cdots(s+n)}{s \cdot n!}.\]
2.2. The product of exponential factors simplifies to:\[\prod_{k=1}^n e^{-s/k} = \exp\left( -s \sum_{k=1}^n \frac{1}{k} \right) = e^{-s H_n}.\]
2.3. Combining these factors yields\[P_n(s) = \frac{s(s+1)\cdots(s+n)}{s \cdot n!} e^{-s H_n}.\]
3. Introducing the sequence \(\gamma_n = H_n - \log n\):Proof:
3.1. Multiply \(P_n(s)\) by \(s e^{s \gamma_n}\), where \(\gamma_n = H_n - \log n\):\[s e^{s \gamma_n} P_n(s) = s e^{s(H_n - \log n)} \cdot \left[ \frac{s(s+1)\cdots(s+n)}{s \cdot n!} e^{-s H_n} \right].\]
3.2. The \(s\) in the numerator and denominator cancel, and the \(e^{s H_n}\) and \(e^{-s H_n}\) cancel:\[s e^{s \gamma_n} P_n(s) = e^{-s \log n} \frac{s(s+1)\cdots(s+n)}{n!} = n^{-s} \frac{s(s+1)\cdots(s+n)}{n!} = \frac{s(s+1)\cdots(s+n)}{n^s n!}.\]
4. Passing to the limit \(n \to \infty\):Proof:
4.1. Since \(\gamma_n \to \gamma\) as \(n \to \infty\), continuity of the exponential function implies \(\lim_{n \to \infty} e^{s \gamma_n} = e^{s \gamma}\).
4.2. Using step 1.3 for \(\lim_{n \to \infty} P_n(s)\):\[\lim_{n \to \infty} \frac{s(s+1)\cdots(s+n)}{n^s n!} = \lim_{n \to \infty} \left[ s e^{s \gamma_n} P_n(s) \right] = s e^{\gamma s} \left( \frac{e^{-\gamma s}}{s \Gamma(s)} \right) = \frac{1}{\Gamma(s)}.\]
4.3. Since \(\Gamma(s) \neq 0\) and \(s \notin \{0, -1, -2, \dots\}\), taking the reciprocal of both sides gives:\[\Gamma(s) = \lim_{n \to \infty} \frac{n^s n!}{s(s+1)(s+2)\cdots(s+n)}.\]