gamma function with negative argument

Question

For $k=0,1,2...$ and small $z$ I want to show that $$\Gamma (-k + z) = \frac{ a_k}{z} + b_k + O(z).$$ I understand that the gamma function cannot be expressed as $$\Gamma ( z) = \int_0 ^\infty e^{-x} x^{z-1} dx$$ for negative values of $z$. The graph below shows that for $z \in \mathbb{z^-}$ the function is alternating between positive and negative infinity thus the need for the small value to be included. Graph of \Gamma(z)

Answer 1

$a_k~=~\lim\limits_{z\to0}~z~\Gamma(z-k)~=~\dfrac{(-1)^k}{k!}$

$b_k~=~\lim\limits_{z\to0}~\Gamma(z-k)-\dfrac{a_k}z~=~-\dfrac{(-1)^k}{k!}\cdot\gamma~-~\dfrac{S_1(k+1,~2)}{k!^2}$

See Euler-Mascheroni constant and Stirling numbers of the first kind for more information.