Poincaré asymptotic series

Remarks

An equivalent condition is that $f(z) - \sum_{n=0}^{N} a_n z^{-n} = O(z^{-N-1})$ for every $N$. The coefficients are determined by $f$: $a_0 = \lim f(z)$ and $a_N = \lim z^N \big( f(z) - \sum_{n<N} a_n z^{-n} \big)$. The same definition with $(z - z_0)^n$ in place of $z^{-n}$ gives Poincaré series at a finite point $z_0$.

Example: $f(z) = \frac{1}{z - 1}$ on $D = (2, \infty)$ has $f(z) \sim \sum_{n \ge 1} z^{-n}$. Check: $f(z) - \sum_{n=1}^{N} z^{-n} = \frac{z^{-N}}{z - 1}$, and $z^N$ times this is $\frac{1}{z - 1} \to 0$.

Non-example: $f(z) = \sin z$ on $D = (0, \infty)$ has no Poincaré asymptotic series, because $a_0 = \lim_{z \to \infty} \sin z$ does not exist.

The function $e^{-z}$ on $(0, \infty)$ has the zero series, like the function $0$, so the series does not determine the function.