Difference between revisions of "Gamma function written as infinite product"

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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$\Gamma(z) = \dfrac{1}{z} \displaystyle\prod_{k=1}^{\infty} \dfrac{(1+\frac{1}{k})^z}{1+\frac{z}{n}},$$
+
$$\Gamma(z) = \dfrac{1}{z} \displaystyle\prod_{k=1}^{\infty} \dfrac{(1+\frac{1}{k})^z}{1+\frac{z}{k}},$$
 
where $\Gamma$ denotes the [[gamma]] function.
 
where $\Gamma$ denotes the [[gamma]] function.
  
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==References==
 
==References==
 
* {{BookReference|A course of modern analysis|1920|Edmund Taylor Whittaker|author2=George Neville Watson|edpage=Third edition|prev=Reciprocal gamma written as an infinite product|next=findme}}: $\S 12 \cdot 11$
 
* {{BookReference|A course of modern analysis|1920|Edmund Taylor Whittaker|author2=George Neville Watson|edpage=Third edition|prev=Reciprocal gamma written as an infinite product|next=findme}}: $\S 12 \cdot 11$
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=Gamma function written as a limit of a factorial, exponential, and a rising factorial|next=Reciprocal gamma written as an infinite product}}: §1.1 (2)
+
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Gamma function written as a limit of a factorial, exponential, and a rising factorial|next=Reciprocal gamma written as an infinite product}}: §1.1 (2)
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 20:56, 3 March 2018

Theorem

The following formula holds: $$\Gamma(z) = \dfrac{1}{z} \displaystyle\prod_{k=1}^{\infty} \dfrac{(1+\frac{1}{k})^z}{1+\frac{z}{k}},$$ where $\Gamma$ denotes the gamma function.

Proof

References