Proof from From Visual Complex Functions From Visual Complex Functions with Phase Portraits by Wegert (author).
From the book. I am stuck on the proof of the theorem.
I have indicated by
( ͡° ͜ʖ ͡°) where in the proof I start to get lost. The definition for $q$ confuses me too.
Here's the theorem that's being proved.
Theorem 3.2.9.  If $f$  is analytic at $z_{0}$ and $f(z_{0}) \neq 0$,  then $1/f$   is analytic at $z_{0}.$
The Taylor coefficients $b_{k}$  of  $1/f$  at $z_{0}$  can be computed recursively from the Taylor coefficients $a_{k}$  of $f$ by $b_{0} :=1/a_{0}$ and $b_{k} :=-\displaystyle \frac{1}{a_{0}}(a_{1}b_{k-1}+a_{2}b_{k-2}+\ldots+a_{k}b_{0})$ , $k=1$, 2, . . . .   (3.37)
The proof makes use of the following lemma.
Lemma 3.2.2 (Abel-Weierstrass).  Let $R$  be the radius of convergence of the power series (3.16).
(i)  If $0\leq r< R$,   then there exists a constant $c$   such that for all $k\in\mathbb{N}$
$r^{k}| a_{k}|\ \leq c$.   (3.22)
(ii)   If there exist positive numbers $r$   and $c$  such that  (3.22)  holds for all sufficiently large $k\in\mathbb{N}$,  then $R\geq r.$
The inequality (3.22) is also known as  Cauchy's estimate.
Please explain from the line to the end.
Here is the Actual steps in the proof:--
Proof. 1. In the first step we assume that the function $1/f$ is analytic at $z_{0}$. Then
its Taylor series
$\displaystyle \frac{1}{f(z)}=b_{0}+b_{1}(z-z_{0})+b_{2}(z-z_{0})^{2}+\ldots+b_{k}(z-z_{0})^{k}+\ldots$   (3.38)
converges in a neighborhood of $z_{0}$ and its Cauchy product with the Taylor series of $f$ is the constant function 1. The latter is equivalent to the infinite system of equations
\begin{aligned}
a_{0}b_{0}=1  \\
a_{0}b_{1}+a_{1}b_{0}=0 \\
a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}=0
\end{aligned}
Since $a_{0}\neq 0$, this triangular system can be solved with respect to the coefficients
$b_{k}$, which yields the recursion (3.37).


*It remains to prove that the series (3.38), with coefficients $b_{k}$ given by the recursion (3.37), indeed has a positive radius of convergence.

By Cauchy's estimate (3.22) in Lemma 3.2.2, there are positive numbers $c$
and $r$ such that $|a_{n}| \leq cr^{-n}$ for all $n\in \mathbb{N}$. We set $q :=1+c/|a_{0}|$ and show that
( ͡° ͜ʖ ͡°) $|b_{n}| \displaystyle \leq\frac{c}{|a_{0}|^{2}}\frac{q^{n-1}}{r^{n}},n=$1, 2, . . . .   (3.39)
For $n=1$ we have $b_{1} =-a_{1}/a_{0}^{2}$ and $|a_{1}| \leq c/r$, so that indeed
$$
|b_{1}|=\frac{a_{1}}{a_{0}^{2}}\leq\ \frac{c}{|a_{0}|^{2}}\frac{1}{r}.
$$
Now we assume that (3.39) holds for all $n=1$, 2, . . . , $k-1$ and consider the case where $n=k$. Using $|b_{0}| = 1/|a_{0}|$, the recursive definition of $b_{k}$, and the triangle inequality, we estimate
$$
|b_{k}|\leq\frac{1}{|a_{0}|}(|a_{k}b_{0}|+\sum_{j=1}^{k-1}|a_{k-j}||b_{j}|)
$$
$$
\leq\frac{1}{|a_{0}|}(|a_{k}b_{0}|+\sum_{j=1}^{k-1}\frac{c}{r^{k-j}}\frac{c}{r^{j}|a_{0}|^{2}}q^{j-1})
$$
$$
\leq\frac{c}{r^{k}|a_{0}|^{2}}(1+\frac{c}{|a_{0}|}\ \sum_{j=0}^{k-2}q^{j})
$$
$$
=\frac{c}{r^{k}|a_{0}|^{2}}(1+\frac{c}{|a_{0}|}\frac{q^{k-1}-1}{q-1})\ =\frac{c}{r^{k}|a_{0}|^{2}}q^{k-1},
$$
which gives (3.39) for $n= k$ and thus for all $n$. Consequently, by Lemma 3.2.2, the power series (3.38) has radius of convergence not less than $r/q. \square $
Since posting I looked hard at it and it has suddenly all opened up to me like a flower bud tighly closed now fully open. I understand.
 A: Rough notes to help main question.
Proof follows two steps.
In step 1. assume $1/f(z_0)$ IS analytic at $ z_0$. It should be as $f({z_0})$ is non zero.
Therefore it has a Taylor series about ${z_0}$, and thus a radius of convergence (R.O.C.) about the point ${z_0}$. The radius of convergence could be zero $({z_0}$ only), infinite (the whole of $ C$) or some where in between. 
In step 2 we show it's indeed the latter by proving it has a positive radius not less than a positive quantity.
All we have to work with is what we know about the function f and its Taylor series which is analytic  and defined at $ {z_0}.$ We also know Cauchy's estimate for the modulus of the  coefficients of the Taylor series.
We WANT to prove 3.39 is true for ALL natural numbers n $ \in  \Bbb N$.
We use proof by induction and this necessitates a starting point which using the simplest use n=1. Then assume TRUE for n  up to and including $k - 1$ and deduce that it implies true for n=k and hence for all natural numbers.
We only have the induction given for $b_n$ and apply the triangle inequality. After peeling off the term $ {a_k}{b_0}$  we are left with a sum which we sum from right-to-left. (I don't think that matters but that is the way in the book).The minus vanishes when the modulus is taken.
The definition for q works near the end when the whole thing simplifies at the inductive step.
Apart from some direct algebra taking constants outside the sum and factorizing the use of a substitution in the sum to simplify the indexed start and end values is done. This can be accomplished by say replacing j by $m := j - 1$. Then calling $m$ $j$ again after (as it's just a dummy index). The finite geometric power-series is dealt with by the commonly known formula noting that there are $k - 1$ terms and it starts with $1$. The final part simplifies by noticing that $c/\left| {{a_0}} \right|$ is $q - 1$ (by definition above). So cancels with the same).
Finally, a last appeal to the second part of the lemma. The index - subscript and the index in the power (of r) is the same number. Instead of using the coefficient a we replace with b, and since $q ^(k-1)$  equals $q^k/q$  replace r by r/q and for the R.H.S. c>0 such that $c:= c/{\left| {{a_0}} \right|^2}$ times $1/q$.  Now by Cauchy's estimate we can deduce that there exists $0 \prec r \leqslant R$ $\text{R}$ is the R.O.C. for the Taylor series of $ 1/f.$ So R.O.C. ($R$) is not less than some positive quantity ($r:=r/q$). 
