The radius of convergence \(r\) is the inverse of the growth ratio of the sequence \(a_n,\) defined as the infimum in \([0,+\infty]\) of the set of values \(\sigma \in [0,+\infty)\) such that \(a_n\) is eventually dominated by \(\sigma^n\): \[ \exists \, m \in \mathbb{N}, \; \forall \, n \in \mathbb{N}, \; (n \geq m) \, \Rightarrow \, |a_{n}| \leq \sigma^{n}. \] (or equivalently, such that \(\exists \, \kappa > 0,\) \(\forall \, n \in \mathbb{N},\) \(|a_{n}| \leq \kappa \sigma^{n}.\)) This growth ratio is equal to \(\limsup_{{n} \to +\infty} |a_{n}|^{1/{n}},\) which leads to the Cauchy-Hadamard formula¹: \[ r = \frac{1}{\displaystyle \limsup_{n \to +\infty} |a_{n}|^{1/{n}}}. \] By convention here, \(1/0 = +\infty\) and \(1/(+\infty) = 0.\)

We now prove that the growth ratio of \(|a_n|\) is equal \(\limsup_n |a_n|^{1/n}.\) Indeed, for any \(\sigma\) greater than the growth ratio \(\rho,\) eventually \(|a_n| \leq \sigma^n,\) hence \(|a_n|^{1/n} \leq \sigma\) and \(\limsup_n |a_n|^{1/n} \leq \sigma,\) therefore \(\limsup_n |a_n|^{1/n} \leq \rho.\) Conversely, if \(\sigma\) is smaller than the growth ratio, there is a strictly increasing sequence of \(n \in \mathbb{N}\) such that \(|a_n| > \sigma^n,\) hence \(|a_n|^{1/n} > \sigma\) and \(\limsup_n |a_n|^{1/n} \geq \sigma,\) thus \(\limsup_n |a_n|^{1/n} \geq \rho.\) ‌\(\blacksquare\)

The growth bound of any polynomial sequence \(a_n\) is at most \(1\): the inequality \[|\alpha_0 + \alpha_1 n + \dots + \alpha_p n^p| \leq \rho^n\] holds for any \(\rho > 1\) eventually. Now, for any nonzero polynomial sequence \(a_n\) and any sequence \(b_n\), eventually \(|b_n|\) is dominated by a multiple of \(|a_n b_n|\), thus the growth bound of \(|b_n|\) is at most the growth bound of \(|a_n b_n|\). Reciprocally, the growth bound of \(|a_n b_n|\) is at most the product of the growth bound of \(|a_n|\) – at most one – and the growth bound of \(|b_n|\) and thus at most the growth bound of \(|b_n|\). ‌\(\blacksquare\)

Proof – Lemma.

as expected. ‌\(\blacksquare\)

Now, let \(z \in D(c, r)\) and let \(\rho \in \left]0,r \right[\) such that \(|z-c| < \rho.\) Cauchy’s integral formula provides \[ f(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} \, dw. \] For any \(w \in \gamma([0,1]),\) we have \[ \frac{1}{w-z} = \frac{1}{(w-c) - (z-c)} = \frac{1}{w-c}\frac{1}{1 - \frac{z-c}{w-c}}. \] Since \[ \left|\frac{z-c}{w-c}\right| = \frac{|z-c|}{\rho} < 1, \] we may expand \(f(w)/(w-z)\) into \[ \frac{f(w)}{w-z} = \frac{f(w)}{w-c}\frac{1}{1 - \frac{z-c}{w-c}} = \sum_{n=0}^{+\infty} \frac{f(w)}{w-c} \left(\frac{z-c}{w-c}\right)^n. \] The term of this series is dominated by \[ \frac{\sup_{|w-c|=\rho} |f(w)|}{\rho} \left(\frac{|z-c|}{\rho}\right)^n; \] the convergence of the series is normal – and thus uniform – with respect to the variable \(w.\) Finally \[ \begin{split} f(z) &= \int_{\gamma} \left[ \sum_{n=0}^{+\infty}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \right] dw \\ &= \sum_{n=0}^{+\infty} \left[ \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \, dw \right]\\ &= \sum_{n=0}^{+\infty} \left[ \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n \end{split} \] which is the desired expansion. ‌\(\blacksquare\)

1. The open annulus \(A(0, 0, +\infty),\) centered on the origin, with inner radius \(0\) and outer radius \(+\infty,\) is the set \(\mathbb{C}^*.\)

2. The sets \(A(0,0,1),\) \(A(0,1,2)\) and \(A(0,2,+\infty)\) are three open annuli centered on the origin and included in the open set \(\Omega = \mathbb{C} \setminus \{i,2\}.\) They are maximal in \(\Omega\) – if we decrease their inner radius and/or increase their outer radius the resulting annulus is not a subset of \(\Omega\) anymore.

Now, for any \(z \in A(c,r_1,r_2),\) there is an open neighbourhood \(U\) of \(z\) where \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is normally convergent and an open neighbourhood \(V\) of \((z-c)^{-1}\) in \(\mathbb{C}^*\) where \(\sum_{n=1}^{+\infty} a_{-n}w^{n}\) is normally convergent. The Laurent series \(\sum_{n=-\infty}^{+\infty} a_n (z-c)^n\) is normally convergent in the open neighbourhood \(U \cap \{w^{-1} + c \; | \; w \in V\}\) of \(z.\) ‌\(\blacksquare\)

Let \(z \in A(c, r_1, r_2)\) and \(\rho_1,\rho_2 \in \left]r_1,r_2\right[\) such that \(\rho_1 <|z-c| < \rho_2.\) Let \(\gamma_1 = c + \rho_1 [\circlearrowleft]\) and \(\gamma_2 = c + \rho_2 [\circlearrowleft];\) Cauchy’s integral formula provides \[ f(z) = \frac{1}{i2\pi} \int_{\gamma_2} \frac{f(w)}{w-z} \, dw - \frac{1}{i2\pi} \int_{\gamma_1} \frac{f(w)}{w-z} \, dw \] As in the proof of “Power Series Expansion”, we can establish that \[ \frac{1}{i2\pi} \int_{\gamma_2} \frac{f(w)}{w-z} \, dw = \sum_{n=0}^{+\infty} \left[ \frac{1}{i2\pi}\int_{\gamma_2}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n. \] A similar argument, based on a series expansion of \[ \frac{1}{w-z} = -\frac{1}{(z-c) - (w-c)} = -\frac{1}{z-c}\frac{1}{1 - \frac{w-c}{z-c}} \] yields \[ \frac{1}{i2\pi} \int_{\gamma_1} \frac{f(w)}{w-z} \, dw = -\sum_{n=-1}^{-\infty} \left[ \frac{1}{i2\pi}\int_{\gamma_1}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n. \] The combination of both expansions provides the expected result. ‌\(\blacksquare\)