Consider the following naive manipulations:
\begin{align} \int_0^\infty \frac{e^{-x}}{1+ax}\:dx & = \int_0^\infty e^{-x}\frac{1}{1-(-ax)}\:dx\\ &= \int_0^\infty e^{-x} \left( \sum_{n=0}^\infty (-ax)^n \right) \: dx\\ &= \sum_{n=0}^\infty (-a)^n \int_0^\infty e^{-x} x^n \: dx\\ &= \sum_{n=0}^\infty n! (-a)^n \\ \end{align}
These are wrong, of course, since the integral we started with exists yet the partial sum $$\sum_{n=0}^m n! (-a)^n$$ is divergent. Yet for small $a$ and $m$ the approximation is quite accurate. Why does it "work" for small $a$ and $m$?