What's the approximation of Gamma(x) when x goes to 0?

Question

I need to find a sequence of $ \alpha_N, \beta_N$ that makes $ \frac{\varGamma(\alpha_N)}{\beta_N}\cdot\frac{\alpha_N}{\beta_N}\big[\frac{1}{n}-\frac{n}{(n+\beta-1)^2}\big] $ goes to $0$. I try to let $\alpha_N, \beta_N$ both goes to $0$ with different rate, but I was stuck by the $\varGamma(\alpha_N)$, can anyone help to give me a approximation for it?

Answer 1

Here it is: as $x\to 0$ we have, up to the fourth order:

$$\Gamma(x) \sim \frac{1}{x}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) x+\frac{1}{6} x^2 \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)+\frac{1}{24} x^3 \left(\gamma ^4+\frac{3 \pi ^4}{20}+\gamma ^2 \pi ^2-4 \gamma \psi ^{(2)}(1)\right)+O\left(x^4\right)$$

Where:

• $\gamma$ is the Euler-Mascheroni constant;

• $\psi^{(2)}(1)$ is the Polygamma Function (the second derivative of the Digamma Function, in this case).

Notice that for very small $x$, you simply have that $\Gamma(x)\approx \frac{1}{x}$. Indeed the Gamma function has a pole at $x = 0$.