The proof of Konig's theorem for complex analysis: my attempt has left me with an indication that the theorem is *not* true - what's going on? I want to show:

Given a meromorphic function $f$ on the disk $|z|\lt R$, with only one singularity, namely a simple pole $\zeta$, in this disk, be written as $f(z)=\sum_{n=0}^\infty c_nz^n$, with the condition $c_0\neq0$. Let $\sigma\in(0,1)$ be such that $|\zeta|\lt\sigma R$. Then: $$\frac{c_n}{c_{n+1}}=\zeta+o(\sigma^{n+1})$$

This is Konig's theorem and I would happily see someone else's proof, but after a frustrated search online it would seem that this theorem has fallen through the cracks a little.
My current attempt, after using different ideas for an unsuccessful while, building off of the attempt given in this question:

By the Laurent expansion of a simple  pole, we have, in any suitably small neighbourhood of $\zeta$, where $r=\operatorname{Res}(f,\zeta)$: $$\begin{align}f(z)&=\frac{r}{z-\zeta}+\sum_{m=0}^\infty b_m(z-\zeta)^m=\frac{r}{z-\zeta}+\sum_{m=0}^\infty b_m\sum_{n=0}^m\binom{m}{n}z^n(-\zeta)^{m-n}\\&=\frac{r}{z-\zeta}+\sum_{n=0}^\infty z^n\cdot\sum_{m=0}^\infty\binom{m}{n}b_n\cdot(-\zeta)^{m-n}\\&\overset{\color{red}*}{=}-\frac{r}{\zeta}\sum_{n=0}^\infty\left(\frac{z}{\zeta}\right)^n+\sum_{n=0}^\infty z^n\cdot\sum_{m=0}^\infty\binom{m+n}{n}b_{m+n}\cdot(-\zeta)^n\end{align}$$ Using the convention $\binom{a}{b}=0\iff a<b$. Comparing coefficients, as the Laurent series must coincide with the theorem-statement Taylor series in some punctured neighbourhood of $\zeta$: $$c_n=-\frac{r}{\zeta^{n+1}}+\psi(n),\,\psi(n)=\sum_{m=0}^\infty\binom{m+n}{n}b_{m+n}\cdot(-\zeta)^n$$
We want to show: $$\frac{c_n}{c_{n+1}}=\frac{-r\zeta^{-(n+1)}+\psi(n)}{-r\zeta^{-(n+2)}+\psi(n+1)}=\zeta+o(\sigma^{n+1})$$ But I can find no way to isolate $\zeta$ from this expression.

The question I linked assumed that $\lim_{n\to\infty}\psi(n)=0$, and thus they believed they showed that $\lim_{n\to\infty}\frac{c_n}{c_{n+1}}=\zeta$. This is also the "intuition" provided by Wikipedia's article, also linked. However, I am specifically interested in the equality with $\zeta+o(\sigma^{n+1})$. I am strongly unconvinced that $\psi(n)\to0$ since some experimentation with WolframAlpha leads me to believe otherwise - also, the binomial coefficients will become huge in the series, as would $\zeta^n$.
The step I marked with a red star was assumed in Wikipedia's "intuition" section, but I have my doubts that one can guarantee $\left|\frac{z}{\zeta}\right|\lt1$, so the geometric series seems dubious to me. Maybe a Taylor expansion would be more suitable - I don't know.
Is this approach doomed to failure, or can one recover the theorem from this? As there is an absence of online mention of this theorem (the top $15$ webpages I found had simply scraped the text from Wikipedia's article...) if anyone knows or knows a link to a proof I would greatly appreciate that!
 A: Expanding $f$ into a series centered at $\zeta$ and then re-expanding about zero is way too complicated. The proof is in fact much easier. If $r$ is the residue of $f$ at $\zeta$, then the function $g(z) = f(z) - \frac{r}{z-\zeta} = \sum_{n=0}^\infty b_n z^n$ is analytic in the disk $|z|<R$, so that (by Cauchy-Hadamard formula for the radius of convergence or similar estimates for the size of coefficients of convergent power series) you have $\limsup |b_n|^{1/n} \le 1/R$. In particular, for any $S$ with $0<S<R$ there exists a constant $C$ such that $|b_n| \le C S^{-n}$ for all $n$. Since $|\zeta| < \sigma R$, we can pick such $S$ with $|\zeta| < \sigma S$. Now we write
$$
f(z) = g(z) + \frac{r}{z-\zeta} = \sum_{n=0}^\infty b_n z^n - r \sum_{n=0}^\infty \frac{z^n}{\zeta^{n+1}} = \sum_{n=0}^\infty \left(b_n - \frac{r}{\zeta^{n+1}} \right) z^n, 
$$
so that $c_n = b_n - r \zeta^{-n-1}$, and thus
$$
\left| \frac{c_n}{c_{n+1}} - \zeta \right| = \left| \frac{b_n-r \zeta^{-n-1}}{b_{n+1} - r \zeta^{-n-2}} - \zeta \right| = \left| \frac{b_{n+1}-\zeta b_n}{b_{n+1} - r \zeta^{-n-2}} \right|.
$$
Using the estimates
$$
|b_{n+1}-\zeta b_n| \le B S^{-n},
$$
which holds with $B = (1+\sigma R)C$, and
$$
|b_{n+1} - r \zeta^{-n-2}| \ge |r| |\zeta|^{-n-2} - |b_{n+1}| \ge |r| |\zeta|^{-n-2} - C S^{-n},
$$
we get that
$$
\left| \frac{c_n}{c_{n+1}} - \zeta \right| \le \frac{B S^{-n}}{|r| |\zeta|^{-n-2} - CS^{-n}} = \frac{B}{A \rho^n - C} = O(\rho^{-n})
$$
with $A = |r| |\zeta|^{-2}$ and $\rho = \frac{S}{|\zeta|}$. By the choice of $S$ above we know that $\rho^{-1} = \frac{|\zeta|}{S} < \sigma$, so that $O(\rho^{-n}) = o(\sigma^n) = o(\sigma^{n+1})$ (not sure why the exponent is $n+1$, not $n$, in the claim, but these are equivalent), proving the claim.
