Taylor expansion at $\infty$

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What does it mean to:

Take a Taylor expansion at infinity

A taylor expansion of $\Gamma (x)$ centered at 0 gives me: $$\frac{1}{x}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) x+\frac{1}{6} x^2 \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)+O\left(x^3\right)$$

But an expansion of $\Gamma (x)$ centered at $\infty$ gives me $$\left(\sqrt{2 \pi } \sqrt{\frac{1}{x}}+\frac{1}{6} \sqrt{\frac{\pi }{2}} \left(\frac{1}{x}\right)^{3/2}+O\left(\left(\frac{1}{x}\right)^{5/2}\right)\right) \exp \left(\left(-\log \left(\frac{1}{x}\right)-1\right) x+O\left(\left(\frac{1}{x}\right)^3\right)\right)$$

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In my experience, when one asks for a power series about $\infty$, what is usually meant is a series about $0$, chosen so that its annulus of convergence borders $\infty$.

For example, consider the function

$$ f(x) = \frac{1}{1-x} + \frac{1}{2-x} $$

This has three different convergent series about $x=0$:

$$f(x) = \sum_{n=0}^{\infty} \left(1 + \frac{1}{2^{n+1}} \right) x^n \qquad \qquad |x| < 1$$ $$f(x) = \sum_{n=-\infty}^{-1} (-1) x^n + \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} x^n \qquad \qquad 1 < |x| < 2$$ $$f(x) = \sum_{n=-\infty}^{-1} (-1)\left( 1 + \frac{1}{2^{n+1}} \right) x^n \qquad \qquad 2 < |x|$$

The last of these converges in a neighborhood of $\infty$, and is most likely what people mean if they ask for a series about $\infty$ for $f$.