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On page 255 of the book "Handbook of Mathematical Functions" by Milton Abramowitz and Irene A. Stegun, it is mentioned that

$$\lim_{z \to n} \frac{1}{\Gamma{(-z)}}=0=\frac{1}{(-n-1)!}.$$

I am not able to understand why it is equal to $\frac{1}{(-n-1)!}$. Thank you in advance for any help.

user103005
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    It equals that because $\Gamma(n) = (n-1)!$ is how we extend the factorial function. It is by definition. Just like $(1/2)!$ and other numbers are, by definition, defined by the Gamma function.

    The limit is a property of the Gamma function, I suppose.

     – user357980 Mar 03 '18 at 03:26
  • The values $\Gamma(0)$, $\Gamma(-1)$, $\Gamma(-2)$, and so on, are all $\infty$. If you use the extension of factorial mentioned by the comment above, it would correspond to taking $(-1)!=\infty$, $(-2)!=\infty$, $(-3)!=\infty$, and so on. But it may be seen as abuse of notation. But do not that the reciprocal gamma function $1/\Gamma$ is defined (finite) in every point; in fact an entire function. – Jeppe Stig Nielsen Mar 03 '18 at 03:34

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