Divergent asymptotic series

Example

For $x > 0$ let $$F(x) = \int_0^\infty \frac{e^{-t}}{1 + x t}\,dt.$$ Then, as $x \to 0^+$, $$F(x) \sim \sum_{n \ge 0} (-1)^n n!\, x^n = 1 - x + 2x^2 - 6x^3 + \cdots,$$ and this series is a divergent series for every $x \ne 0$.

Check of the expansion: for $N \ge 0$ and $t \ge 0$, $$\frac{1}{1 + x t} = \sum_{n=0}^{N} (-x t)^n + \frac{(-x t)^{N+1}}{1 + x t}.$$ Integrating against $e^{-t}$ and using $\int_0^\infty t^n e^{-t}\,dt = n!$ gives $$F(x) - \sum_{n=0}^{N} (-1)^n n!\, x^n = R_N(x), \qquad \abs{R_N(x)} \le x^{N+1} \int_0^\infty t^{N+1} e^{-t}\,dt = (N + 1)!\, x^{N+1}.$$ So the error is $O(x^{N+1})$ for each $N$, which is the condition of a Poincaré asymptotic series at $0$.

Check of divergence: for $x \ne 0$ the ratio of consecutive terms is $(n + 1) \abs{x} \to \infty$, so the terms do not tend to $0$.