[proofplan] We decompose the bi-infinite series into two ordinary power series: the "analytic part" $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ and the "principal part" $\sum_{n=1}^{\infty} a_{-n} (z - z_0)^{-n}$. We analyse each using the Cauchy–Hadamard formula (the [Root Test](/theorems/175)). The analytic part converges absolutely for $|z - z_0| < R$ and diverges for $|z - z_0| > R$. The principal part, after substituting $w = 1/(z - z_0)$, converges absolutely for $|z - z_0| > r$ and diverges for $|z - z_0| < r$. Both parts converge absolutely on the annulus $r < |z - z_0| < R$, and the [Weierstrass M-Test](/theorems/261) upgrades this to uniform convergence on compact subsets. [/proofplan] [step:Decompose the bi-infinite series into analytic and principal parts] Write the bi-infinite series as \begin{align*} \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n = \underbrace{\sum_{n=0}^{\infty} a_n (z - z_0)^n}_{=: S_+(z)} + \underbrace{\sum_{n=1}^{\infty} a_{-n} (z - z_0)^{-n}}_{=: S_-(z)}. \end{align*} Define the analytic part $S_+: \mathbb{C} \to \mathbb{C}$ as the standard power series in $z - z_0$, and the principal part $S_-: \mathbb{C} \setminus \{z_0\} \to \mathbb{C}$ as the power series in $(z - z_0)^{-1}$. The bi-infinite series converges absolutely at a point $z$ if and only if both $S_+(z)$ and $S_-(z)$ converge absolutely. [guided] A bi-infinite series $\sum_{n=-\infty}^{\infty} a_n (z - z_0)^n$ is, by definition, the sum of two independent series: the non-negative powers and the negative powers. We write \begin{align*} \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n = \underbrace{\sum_{n=0}^{\infty} a_n (z - z_0)^n}_{=: S_+(z)} + \underbrace{\sum_{n=1}^{\infty} a_{-n} (z - z_0)^{-n}}_{=: S_-(z)}. \end{align*} We define the analytic part $S_+: \mathbb{C} \to \mathbb{C}$ as the standard power series in $z - z_0$, and the principal part $S_-: \mathbb{C} \setminus \{z_0\} \to \mathbb{C}$ as the power series in $(z - z_0)^{-1}$. Since absolute convergence of a sum is equivalent to absolute convergence of each summand (provided both are well-defined), the bi-infinite series converges absolutely at $z$ if and only if both $S_+(z)$ and $S_-(z)$ converge absolutely. This reduction allows us to apply the classical Cauchy–Hadamard theory to each part separately. [/guided] [/step] [step:Apply the Cauchy–Hadamard formula to the analytic part $S_+$] The series $S_+(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n$ is a standard power series. By the Cauchy–Hadamard formula (the [Root Test](/theorems/175) applied to the power series), the radius of convergence is \begin{align*} R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}. \end{align*} The series $S_+$ converges absolutely for $|z - z_0| < R$ and diverges for $|z - z_0| > R$. [guided] The series $S_+(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n$ is a standard power series in the variable $z - z_0$. The Cauchy–Hadamard formula determines its radius of convergence: applying the [Root Test](/theorems/175) to the sequence $b_n = a_n (z - z_0)^n$, we compute \begin{align*} \limsup_{n \to \infty} |b_n|^{1/n} = \limsup_{n \to \infty} |a_n|^{1/n} \cdot |z - z_0|. \end{align*} The Root Test guarantees absolute convergence when this is strictly less than $1$, i.e., when $|z - z_0| < 1/\limsup_{n \to \infty} |a_n|^{1/n} =: R$, and divergence when this exceeds $1$, i.e., when $|z - z_0| > R$. This is exactly the Cauchy–Hadamard radius of convergence for the analytic part. [/guided] [/step] [step:Reduce the principal part $S_-$ to a power series via the substitution $w = (z - z_0)^{-1}$] Define $w := (z - z_0)^{-1}$. Then \begin{align*} S_-(z) = \sum_{n=1}^{\infty} a_{-n} (z - z_0)^{-n} = \sum_{n=1}^{\infty} a_{-n} w^n =: \tilde{S}(w). \end{align*} The series $\tilde{S}: \mathbb{C} \to \mathbb{C}$ is a standard power series in $w$ with coefficients $a_{-n}$. By the Cauchy–Hadamard formula, $\tilde{S}(w)$ converges absolutely for \begin{align*} |w| < \frac{1}{\limsup_{n \to \infty} |a_{-n}|^{1/n}} = \frac{1}{r}, \end{align*} where $r := \limsup_{n \to \infty} |a_{-n}|^{1/n}$, and diverges for $|w| > 1/r$. Since $|w| = 1/|z - z_0|$, the condition $|w| < 1/r$ translates to $|z - z_0| > r$, and $|w| > 1/r$ translates to $|z - z_0| < r$. Therefore $S_-(z)$ converges absolutely for $|z - z_0| > r$ and diverges for $|z - z_0| < r$. [guided] The principal part $S_-(z) = \sum_{n=1}^{\infty} a_{-n} (z - z_0)^{-n}$ is not a standard power series in $z - z_0$, so we cannot apply the Cauchy–Hadamard formula directly. The key idea is to introduce the substitution $w := (z - z_0)^{-1}$, which converts negative powers of $z - z_0$ into positive powers of $w$: \begin{align*} S_-(z) = \sum_{n=1}^{\infty} a_{-n} (z - z_0)^{-n} = \sum_{n=1}^{\infty} a_{-n} w^n =: \tilde{S}(w). \end{align*} Now $\tilde{S}: \mathbb{C} \to \mathbb{C}$ is a standard power series in $w$ with coefficients $a_{-n}$. By the [Root Test](/theorems/175), the Cauchy–Hadamard radius of convergence of $\tilde{S}$ is \begin{align*} \frac{1}{\limsup_{n \to \infty} |a_{-n}|^{1/n}} = \frac{1}{r}. \end{align*} So $\tilde{S}(w)$ converges absolutely for $|w| < 1/r$ and diverges for $|w| > 1/r$. We now translate back to the $z$-variable. Since $w = (z - z_0)^{-1}$, we have $|w| = 1/|z - z_0|$. Therefore: - $|w| < 1/r$ if and only if $|z - z_0| > r$: absolute convergence of $S_-$. - $|w| > 1/r$ if and only if $|z - z_0| < r$: divergence of $S_-$. The inner radius $r$ of the annulus is thus the "radius of divergence" of the principal part, playing the same role for the negative powers that $R$ plays for the positive powers, but in the reversed direction. [/guided] [/step] [step:Combine: absolute convergence on $A(z_0; r, R)$ and divergence outside] Since $r < R$ by hypothesis, the annulus $A(z_0; r, R) := \{z \in \mathbb{C} : r < |z - z_0| < R\}$ is non-empty. For $z \in A(z_0; r, R)$, both $|z - z_0| < R$ (so $S_+$ converges absolutely) and $|z - z_0| > r$ (so $S_-$ converges absolutely). Therefore the bi-infinite series converges absolutely on $A(z_0; r, R)$. For $|z - z_0| > R$, the analytic part $S_+$ diverges, so the bi-infinite series diverges. For $|z - z_0| < r$, the principal part $S_-$ diverges, so the bi-infinite series diverges. [/step] [step:Upgrade to uniform convergence on compact subsets via the Weierstrass M-Test] Let $K \subset A(z_0; r, R)$ be compact. Since $K$ is a compact subset of the open annulus, there exist $r < r_1 \le r_2 < R$ such that $r_1 \le |z - z_0| \le r_2$ for all $z \in K$. For the analytic part, fix $\rho := r_2 < R$. For all $z \in K$ and $n \ge 0$, \begin{align*} |a_n (z - z_0)^n| \le |a_n| \rho^n. \end{align*} Since $\rho < R = 1/\limsup_{n \to \infty} |a_n|^{1/n}$, we have $\limsup_{n \to \infty} (|a_n| \rho^n)^{1/n} = \rho \cdot \limsup_{n \to \infty} |a_n|^{1/n} < 1$, so $\sum_{n=0}^{\infty} |a_n| \rho^n$ converges by the [Root Test](/theorems/175). By the [Weierstrass M-Test](/theorems/261) with $M_n := |a_n| \rho^n$, the series $S_+$ converges uniformly on $K$. For the principal part, fix $\sigma := 1/r_1$. Since $|z - z_0| \ge r_1$ on $K$, we have $|(z - z_0)^{-1}| \le \sigma$ for all $z \in K$. Thus for $n \ge 1$, \begin{align*} |a_{-n} (z - z_0)^{-n}| \le |a_{-n}| \sigma^n. \end{align*} Since $\sigma = 1/r_1 < 1/r$, the series $\sum_{n=1}^{\infty} |a_{-n}| \sigma^n$ converges (it is the series $\tilde{S}$ evaluated at $|w| = \sigma < 1/r$). By the [Weierstrass M-Test](/theorems/261), the series $S_-$ converges uniformly on $K$. Since both $S_+$ and $S_-$ converge uniformly on $K$, the bi-infinite series $\sum_{n=-\infty}^{\infty} a_n (z - z_0)^n$ converges uniformly on $K$. [guided] We must upgrade absolute convergence to uniform convergence on compact subsets. Let $K \subset A(z_0; r, R)$ be compact. The key observation is that compact subsets of the open annulus stay strictly away from both boundary circles: since the continuous function $z \mapsto |z - z_0|$ attains its minimum and maximum on $K$, there exist constants $r_1, r_2$ with $r < r_1 \le r_2 < R$ such that $r_1 \le |z - z_0| \le r_2$ for all $z \in K$. **Analytic part.** Fix $\rho := r_2 < R$. For all $z \in K$ and $n \ge 0$, we have $|z - z_0| \le \rho$, so \begin{align*} |a_n (z - z_0)^n| \le |a_n| \rho^n =: M_n^+. \end{align*} We need $\sum_{n=0}^{\infty} M_n^+ = \sum_{n=0}^{\infty} |a_n| \rho^n < \infty$. By the [Root Test](/theorems/175), $\limsup_{n \to \infty} (|a_n| \rho^n)^{1/n} = \rho \cdot \limsup_{n \to \infty} |a_n|^{1/n} = \rho/R < 1$ since $\rho < R$. So the majorant series converges, and the [Weierstrass M-Test](/theorems/261) gives uniform convergence of $S_+$ on $K$. **Principal part.** Fix $\sigma := 1/r_1$. For all $z \in K$, we have $|z - z_0| \ge r_1 > r$, so $|(z - z_0)^{-1}| \le 1/r_1 = \sigma$. Therefore for $n \ge 1$: \begin{align*} |a_{-n} (z - z_0)^{-n}| \le |a_{-n}| \sigma^n =: M_n^-. \end{align*} We need $\sum_{n=1}^{\infty} M_n^- = \sum_{n=1}^{\infty} |a_{-n}| \sigma^n < \infty$. Again by the [Root Test](/theorems/175), $\limsup_{n \to \infty} (|a_{-n}| \sigma^n)^{1/n} = \sigma \cdot \limsup_{n \to \infty} |a_{-n}|^{1/n} = \sigma \cdot r = r/r_1 < 1$ since $r_1 > r$. The majorant converges, so by the [Weierstrass M-Test](/theorems/261), $S_-$ converges uniformly on $K$. Since both $S_+$ and $S_-$ converge uniformly on $K$, their sum — the full bi-infinite series — converges uniformly on $K$. As $K$ was an arbitrary compact subset of $A(z_0; r, R)$, the series converges uniformly on compact subsets of the annulus. [/guided] [/step]