Weierstrass's Infinite Product for the Gamma Function
Theorem
The following holds for the Gamma function $\Gamma : (0, \infty) \to \mathbb{R}$: $$ {1 \over \Gamma (x)} = x e^{\gamma x } \lim_{n \to \infty} \prod_{k=1}^{n} \left( 1 + {x \over k} \right) e^{- {x \over k} } $$
- $\gamma$ is the Euler-Mascheroni constant.
Explanation
$$ \Gamma (x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt $$ The Gamma function is defined as above, and also by Euler’s limit formula, $$ \Gamma (x) = \lim_{n \to \infty} {{n^x n!} \over {x(x+1)(x+2) \cdots (x+n) }} $$ Another form of the Gamma function is Weierstrass’s infinite product. By learning this, we come to know the three most famous forms of the Gamma function.
Proof
Theorem: The proof could be shorter if there were more lemmas, but they were not used unnecessarily. The techniques used in this proof are valid here only since learning and familiarizing oneself with lemmas immediately can be difficult. If one has a slight understanding of analysis, understanding this proof should not be difficult.
By Euler’s limit formula $$ \begin{align*} {1 \over \Gamma (x)} =& \lim_{n \to \infty} { {x(x+1)(x+2) \cdots (x+n) } \over {n^x n!} } \\ =& \lim_{n \to \infty} x n^{-x} \left( { {1+x} \over {1} } \right) \left( { {2+x} \over {2} } \right) \cdots \left( { {n+x} \over {n} } \right) \\ =& \lim_{n \to \infty} x n^{-x} \prod_{k=1}^{n} \left( 1 + { x \over k } \right) \\ =& x \lim_{n \to \infty} e^{-x \log{n} } \cdot e^{x \left( {1 \over 1} + {1 \over 2} + \cdots {1 \over n} \right) } \cdot e^{-x \left( {1 \over 1} + {1 \over 2} + \cdots {1 \over n} \right) } \cdot \prod_{k=1}^{n} \left( 1 + { x \over k } \right) \\ =& x \lim_{n \to \infty} e^{ x \left( {1 \over 1} + {1 \over 2} + \cdots {1 \over n} - \log{n} \right) } \prod_{k=1}^{n} \left( 1 + { x \over k } \right) e^{- { x \over k} } \end{align*} $$ Since the Euler-Mascheroni constant $\displaystyle \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^{n} \left( { 1 \over k } \right) - \ln{n} \right)$ exists, showing that $\displaystyle \prod_{n=1}^{\infty} \left( 1 + { x \over n } \right) e^{- { x \over n} }$ exists concludes the proof.
For a suitably large natural number $n$, since $$ \begin{align*} \lim_{n \to \infty} {{{x \over n} - \ln \left( 1 + {x \over n} \right)} \over {x^2 \over 2n^2}} =& \lim_{n \to \infty} {{ {1 \over n} - {1 \over n} { 1 \over {1 + {x \over n}}} } \over {x \over n^2}} \\ =& \lim_{n \to \infty} {{ 1 - { 1 \over {1 + {x \over n}}} } \over {x \over n}} \\ =& \lim_{n \to \infty} n {{ {1 + {x \over n}} - 1 } \over x \left( {1 + {x \over n} } \right)} \\ =& \lim_{n \to \infty} {1 \over {1 + {x \over n} }} \\ =& 1 \end{align*} $$ By the $p$-series test, $\displaystyle \sum_{n=1}^{\infty}{x^2 \over 2n^2}$ converges, and by the limit comparison test, $\displaystyle \sum_{n=1}^{\infty} \left\{{x \over n} - \ln \left( 1 + {x \over n} \right) \right\}$ also converges. Meanwhile, $$ \prod_{k=1}^{\infty} \left( 1 + {x \over k} \right) e^{- {x \over k} } = \displaystyle \exp \left( - \sum_{n=1}^{\infty} \left\{ {x \over n} - \ln \left( 1 + {x \over n} \right) \right\} \right) $$ Therefore, $\displaystyle \prod_{k=1}^{\infty} \left( 1 + {x \over k} \right) e^{- {x \over k} }$ also converges.
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