In the below post, in the first answer(by "Start wearing purple" user) How does one get the Bernoulli numbers via the generating function?
$\begin{align}\frac{1}{1+\frac{x}{2!}+\frac{x^2}{3!}+\ldots}\end{align}$ is written as $\begin{align}1-\left(\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\ldots\right)+\left(\frac{x}{2!}+\frac{x^2}{3!}+\ldots\right)^2-\left(\frac{x}{2!}+\ldots\right)^3+\ldots \end{align}$
But expression of the form $\begin{align}\frac{1}{1+x}\end{align}$ can be expanded into $\begin{align}1-x+x^2+...\end{align}$ only if $\begin{align}|x|<1\end{align}$ but $\begin{align}{\frac{x}{2!}+\frac{x^2}{3!}+\ldots}>=1\end{align}$ So how is it possible to write the above expansion?
The idea behind generating functions consists in placing the terms of a sequence in a Taylor expansion in order to get them back through differentiation. Indeed, $$ \frac{x}{e^x-1} = \sum_{n\ge0} B_n\frac{x^n}{n!} \quad\implies\quad B_n = \left.\frac{\mathrm{d}^n}{\mathrm{d}x^n}\right|_{x=0} \,\frac{x}{e^x-1} $$ In consequence, we only need a function defined on a small neighbourhood around $x=0$, that is why you can assume $|x| \ll 1$ arbitrarily small and $\left|\frac{x}{2!} + \frac{x^2}{3!} + \ldots\right| < 1$ in your expansion.