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Prove that $\Gamma (-n+x)=\frac{(-1)^n}{n!}\left [ \frac{1}{x}-\gamma +\sum_{k=1}^{n}k^{-1}+O(x) \right ]$ I don't know how to do this ? Note that $\gamma $ is the Euler-Mascheroni constant

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Start with following infinite product definition of gamma function by Weierstrass: $$\Gamma(z) = \frac{e^{-\gamma z}}{z}\prod_{k=1}^{\infty}(1+\frac{z}{k})^{-1} e^{\frac{z}{k}}$$ you can prove your assertion for the case $n = 0$. You can then use the functional identities: $$\Gamma(z) = (z-1)\Gamma(z-1) =\cdots= (z-1)\cdots(z-n) \Gamma(z-n)$$ to reduce all the remaining cases to the case $n = 0$. –  achille hui Feb 13 '13 at 14:41
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A standard trick is to use the reflection identity $$\Gamma(-n+x) \Gamma(1+n-x) = -\frac{\pi}{\sin(\pi n - \pi x)}$$ giving, under the assumption of $n\in \mathbb{Z}$ $$ \Gamma(-n+x) = (-1)^n \frac{\pi}{\sin(\pi x)} \frac{1}{\Gamma(n+1-x)} = (-1)^n \frac{\pi}{\sin(\pi x)} \frac{1}{\color\green{n!}} \frac{\color\green{\Gamma(n+1)}}{\Gamma(n+1-x)} $$ Assuming $n \geqslant 0$, $$ \begin{eqnarray}\frac{\Gamma(n+1)}{\Gamma(n+1-x)} &=& \frac{1}{\Gamma(1-x)} \prod_{k=1}^n \frac{1}{1-x/k} \\ &=& \left(1+\psi(1) x + \mathcal{o}(x)\right) \left(1+\sum_{k=1}^n \frac{x}{k} + \mathcal{o}(x) \right) \\ &=& 1 + \left( \psi(1) + \sum_{k=1}^n \frac{1}{k} \right) x + \mathcal{o}(x) \end{eqnarray}$$ where $\psi(x)$ is the digamma function. Also using $$ \frac{\pi}{\sin(\pi x)} = \frac{1}{x} + \frac{\pi^2}{6} x + \mathcal{o}(x) $$ and multiplying we get $$ \Gamma(n+1) = \frac{(-1)^n}{n!} \left( \frac{1}{x} + \psi(1) + \sum_{k=1}^n \frac{1}{k} + \mathcal{O}(x) \right) $$ Further $\psi(1) = -\gamma$, where $\gamma$ is the Euler-Mascheroni constant, arriving at your result.

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What's the downvote for? –  Pedro Tamaroff Feb 14 '13 at 14:09
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