$(1-O(z^n))^{-1}=1+O(z^n)$ Using big-$O$ notation, I wish to show that $\big(1-O(z^n)\big)^{-1}=1+O(z^n)$ for every $n \in \{1,2,\dots\}$, $z$ being a complex variable. The rest of this post is dedicated to making this statement precise.
Preliminary notation
Denote by $\mathbf{D}$ the set of domains $D$ in $\mathbb{C}$ (a domain being a non-empty, open, connected subset of $\mathbb{C}$) that include $0$ (i.e. $0 \in D$).
Denote by $A$ the set of all non-constant, analytic functions $f:D\rightarrow\mathbb{C}$ for some $D\in\mathbf{D}$.
For every $f \in A$ denote by $a^f_0, a^f_1, \dots$ the coefficients of the Taylor expansion of $f$ about $0$, and define $c_f$ to be the first index $n \in \{0,1,2,\dots\}$ such that $a^f_n\neq0$. (Since $f$ is not constant, and therefore in particular not identically zero, $c_f$ is well-defined.)
My question
Let $f\in A$ be such that $c_f > 0$ and such that $f(z) \neq 1$ for every $z$ in $f$'s domain. Define $g := (1-f)^{-1} - 1$. I wish to show that $c_g \geq c_f$.
 A: Your statement, as you state it, is not correct because $g(0) \neq 0$ which means that $c_g = 0$. I suppose what you want to show is $c_{g - 1} \geq c_f$.
It would be easier to define $\mathcal A$ as the union of your $A$ with all constant functions. The set $\mathcal A$ is then a ring under addition and multiplication of functions.
There is an injective homomorphism of rings from $\mathcal A$ to the ring of formal power series $\Bbb C[[Z]]$, which sends every function $f$ to its Taylor expansion at $z = 0$.
What you want to show then follows from the corresponding result in $\Bbb C[[Z]]$, namely for any $f$ with $c_f > 0$ we have $g = (1 - f)^{-1} = 1 + f + f^2 + \cdots$ which satisfies $c_g \geq c_f$ because $g - 1 = f(1 + f + \cdots)$ is a multiple of $f$.
A: I have accepted WhatsUp's answer, but I would like to write a version of WhatsUp's answer that doesn't require redefining $A$ as $\mathcal{A}$, and that avoids formal power series and the homomorphism used in WhatsUp's answer.
Since $f$ is analytic by assumption, it is continuous, and in particular continuous at $0$. Since $c_f > 0$, $f(0) = 0$. We may therefore choose some $r \in (0,\infty)$ such that the open disc $\Delta$ about $0$ of radius $r$ is contained inside $f$'s domain, and such that $\big|f(z)\big| < \frac{1}{2}$ for every $z \in \Delta$. (The number $\frac{1}{2}$ could be replaced by any number in $(0,1)$ without affecting the rest of the proof.)
For every $z \in \Delta$ $\big|f(z)\big| < 1$, and therefore the sequence $\big(f^0(z),f^1(z),\dots\big)$ is summable, and, defining $h:\Delta\rightarrow\mathbb{C}$ as $h(z) := \sum_{n=0}^\infty f^n(z)$, we have, for every $z \in \Delta$,
$$
\begin{align*}
g(z) &= \big(1-f(z)\big)^{-1} - 1\\
&= \big(\sum_{n=0}^\infty f^n(z)\big) - 1\\
&= \sum_{n=1}^\infty f^n(z)\\
&= f(z)\sum_{n=1}^\infty f^{n-1}(z)\\
&= f(z)\sum_{n=0}^\infty f^n(z)\\
&= f(z)h(z)\\
&= \big(\sum_{n=c_f}^\infty a_nz^n\big)h(z).
\end{align*}
$$
Since a power series expansion of $g$ about $0$, if it exists, is unique, it suffices, by Mertens' theorem for the product of infinite series, to prove that $h$ can be represented as a power series about $0$ (recall that a power series that converges in an open disc, converges absolutely there), and to do so, it suffices to prove that $h$ is analytic.
Writing $F := f\big|_\Delta$ (i.e. $F$ is the restriction of $f$ to $\Delta$), we note that the function sequence $\big(F^0, F^1,\dots\big)$ is summable pointwise, and that $h = \sum_{n=0}^\infty F^n$. Furthermore, $\big|F^n(z)\big|\leq \frac{1}{2^n}$ for every $z \in \Delta$, and the sequence $\big(\frac{1}{2^0}, \frac{1}{2^1}, \frac{1}{2^2}, \dots\big)$ is summable. Therefore, by the Weierstrass M-test the sequence $\big(F^0, F^1,\dots\big)$ is summable uniformly. Therefore, since for every $n\in\{0,1,2,\dots\}$ $F^n$ is analytic, so is $h = \sum_{n=0}^\infty F^n$.
