I have this book Calculus, Ninth Edition by Varberg, Purcell, and Rigdon; there's a particular point of a theorem (and another line after that) about Infinite Series that I really don't understand. I quote from Section 9.7, Operations on Power Series:
Let $$f(x) = \sum_{n=0}^{\infty} a_n \cdot x^n$$ and $$g(x) =\sum_{n=0}^{\infty} b_n x^n$$, with both of these series converging at least for $|x| < r$. If the operations of addition, subtraction, and multiplication are performed on these series as if they were polynomials, the resulting series will converge for $|x| < r$ and represent $f(x) + g(x)$, $f(x) - g(x)$, and $f(x) \cdot g(x)$, respectively. If $b_0 \neq 0$ the corresponding result holds for division, but we can guarantee its validity only for $|x|$ sufficiently small.
...
We mention that the operation of substituting one power series into another is also legitimate for $|x|$ sufficiently small, provided that the constant term of the substituted series is zero.
What does it mean by "sufficiently small"? How do I find the limit of how large $|x|$ can be before divisions and substitutions between power series become invalid? My book doesn't go into detail on the exact nature of this restriction and doesn't include a proof of the theorem. If I were dividing one power series by another I don't really know what I should be looking for. I wasn't able to find information about convergence sets for a series generated by two power series divided by each other.
The problem with division is that there may be points where $g(x) = 0$ inside the region of convergence.
If we assume that $b_0 \neq 0$, this means that $g(0) \neq 0$, and since $g$ is continuous, it also follows that $g(x) \neq 0$ for $x$ sufficiently small, say for $|x| < \rho$, which (with some work) shows that division of the power series can be done on $|x|<\rho$.
However, the value of $\rho$ depends on the choice of $g$, and in general there is no simple way to compute it. Take for example $$ g_n(x) = 1 - nx + 0 + 0 + \cdots $$ Then $g_n(\frac1n) = 0$, and if we want to divide the power series of $f$ with the power series of $g_n$, we can only do so for $|x| < \frac1n$ which can be arbitrarily small.