[proofplan] We define the set $S$ of radii at which the [power series](/page/Power%20Series) converges absolutely, prove a key lemma that convergence at one point forces absolute convergence at all closer points via comparison with a geometric series, then take $R = \sup S$ and verify the trichotomy: absolute convergence for $|x - c| < R$, divergence for $|x - c| > R$, and classification into the three cases $R = 0$, $R = \infty$, $0 < R < \infty$. [/proofplan] [step:Define the set of absolute-convergence radii] Define the set \begin{align*} S := \{ r \geq 0 : \text{the series } \sum_{n=0}^{\infty} |a_n| r^n \text{ converges} \}. \end{align*} Since every term of $\sum_{n=0}^{\infty} |a_n| \cdot 0^n$ vanishes for $n \geq 1$, the series converges to $|a_0|$, so $0 \in S$ and $S \neq \varnothing$. [/step] [step:Establish that convergence at one point forces absolute convergence at all interior points] [claim:Boundedness Lemma] Suppose the series $\sum_{n=0}^{\infty} a_n (x_0 - c)^n$ converges for some $x_0 \in \mathbb{R}$ with $x_0 \neq c$. Then for every $x \in \mathbb{R}$ with $|x - c| < |x_0 - c|$, the series $\sum_{n=0}^{\infty} a_n (x - c)^n$ converges absolutely. [/claim] [proof] Since $\sum_{n=0}^{\infty} a_n (x_0 - c)^n$ converges, its general term tends to zero: $a_n(x_0 - c)^n \to 0$ as $n \to \infty$. In particular, the sequence $(a_n(x_0 - c)^n)_{n=0}^{\infty}$ is bounded, so there exists $M > 0$ such that \begin{align*} |a_n (x_0 - c)^n| \leq M \quad \text{for all } n \geq 0. \end{align*} Now fix $x \in \mathbb{R}$ with $|x - c| < |x_0 - c|$, and define the ratio \begin{align*} \rho := \frac{|x - c|}{|x_0 - c|} \in [0, 1). \end{align*} For each $n \geq 0$, factor the general term as \begin{align*} |a_n (x - c)^n| &= |a_n (x_0 - c)^n| \cdot \left(\frac{|x - c|}{|x_0 - c|}\right)^n \leq M \rho^n. \end{align*} Since $0 \leq \rho < 1$, the geometric series $\sum_{n=0}^{\infty} M \rho^n = \frac{M}{1 - \rho}$ converges. Both sequences $(|a_n(x-c)^n|)_{n=0}^{\infty}$ and $(M\rho^n)_{n=0}^{\infty}$ are non-negative with $|a_n(x-c)^n| \leq M\rho^n$ for all $n \geq 0$, so the [Comparison Test](/theorems/173) yields convergence of $\sum_{n=0}^{\infty} |a_n(x-c)^n|$. That is, $\sum_{n=0}^{\infty} a_n(x-c)^n$ converges absolutely. [/proof] [guided] The key question is: why should convergence at a single point $x_0$ propagate to all points closer to the center? The answer rests on a comparison with a geometric series, which is the simplest absolutely convergent series. Since $\sum_{n=0}^{\infty} a_n(x_0 - c)^n$ converges, the general term must satisfy $a_n(x_0 - c)^n \to 0$ as $n \to \infty$ (this is the necessary condition for convergence of a series: if the series converges, then its general term tends to zero). A convergent-to-zero sequence is bounded, so there exists $M > 0$ with \begin{align*} |a_n(x_0-c)^n| \leq M \quad \text{for all } n \geq 0. \end{align*} Now take any $x$ with $|x - c| < |x_0 - c|$ and define $\rho := |x-c|/|x_0-c|$. The strict inequality $|x-c| < |x_0-c|$ ensures $\rho \in [0,1)$. We factor: \begin{align*} |a_n(x-c)^n| &= |a_n| \cdot |x-c|^n = |a_n| \cdot |x_0 - c|^n \cdot \frac{|x-c|^n}{|x_0 - c|^n} \\ &= |a_n(x_0-c)^n| \cdot \rho^n \leq M\rho^n. \end{align*} The geometric series $\sum_{n=0}^{\infty} M\rho^n$ converges because $0 \leq \rho < 1$ (its sum equals $M/(1-\rho)$). Both $(|a_n(x-c)^n|)_{n=0}^{\infty}$ and $(M\rho^n)_{n=0}^{\infty}$ are sequences of non-negative real numbers satisfying $|a_n(x-c)^n| \leq M\rho^n$ for all $n \geq 0$, so the hypotheses of the [Comparison Test](/theorems/173) are met. The Comparison Test gives convergence of $\sum_{n=0}^{\infty} |a_n(x-c)^n|$. Notice the argument gives *absolute* convergence, which is stronger than conditional convergence. This is a distinctive feature of power series: inside the [radius of convergence](/theorems/265), the convergence is always absolute. [/guided] [/step] [step:Take the supremum of $S$ to define $R$ and establish absolute convergence for $|x - c| < R$] Define \begin{align*} R := \sup S \in [0, \infty], \end{align*} where the supremum is taken in the extended [real numbers](/page/Real%20Numbers) (so $R = \infty$ if $S$ is unbounded). We show that $\sum_{n=0}^{\infty} a_n(x - c)^n$ converges absolutely whenever $|x - c| < R$. Fix $x$ with $|x - c| < R$. By the definition of supremum, there exists $r \in S$ with $|x - c| < r \leq R$. Since $r \in S$, the series $\sum_{n=0}^{\infty} |a_n| r^n$ converges, which means $\sum_{n=0}^{\infty} a_n(r)^n$ converges when we substitute $|x_0 - c| = r$ in the Boundedness Lemma (taking $x_0$ to be any point with $|x_0 - c| = r$, e.g., $x_0 = c + r$). Since $|x - c| < r = |x_0 - c|$, the Boundedness Lemma gives absolute convergence of $\sum_{n=0}^{\infty} a_n(x - c)^n$. [guided] Why does the supremum give us the right threshold? The set $S$ collects all radii $r \ge 0$ at which the series $\sum_{n=0}^{\infty} |a_n| r^n$ converges. The Boundedness Lemma guarantees that $S$ is "downward closed": if $r \in S$ and $0 \le r' < r$, then $r' \in S$. This is because $|a_n|(r')^n \le |a_n| r^n$ for all $n$ (since $r' < r$), so convergence of $\sum |a_n| r^n$ implies convergence of $\sum |a_n|(r')^n$ by the [Comparison Test](/theorems/173). Therefore $S$ is an interval of the form $[0, R)$ or $[0, R]$ for some $R \in [0, \infty]$. We define $R := \sup S$. Now fix $x$ with $|x - c| < R$. Since $|x - c|$ is strictly less than the supremum $R$, the characterisation of the supremum provides $r \in S$ with \begin{align*} |x - c| < r \le R. \end{align*} Since $r \in S$, the series $\sum_{n=0}^{\infty} |a_n| r^n$ converges. Taking $x_0 = c + r$ (so $|x_0 - c| = r$), the series $\sum_{n=0}^{\infty} a_n (x_0 - c)^n$ converges. The Boundedness Lemma, applied with this $x_0$ and the given $x$, yields absolute convergence of $\sum_{n=0}^{\infty} a_n(x - c)^n$ because $|x - c| < r = |x_0 - c|$. [/guided] [/step] [step:Establish divergence for $|x - c| > R$] Suppose $|x - c| > R$ and, for contradiction, that $\sum_{n=0}^{\infty} a_n(x - c)^n$ converges. By the Boundedness Lemma, the series converges absolutely at every $y$ with $|y - c| < |x - c|$. In particular, taking $y = c + r'$ for any $r' \in (R, |x - c|)$, we obtain $r' \in S$. But $r' > R = \sup S$, contradicting the definition of the supremum. Therefore the series diverges whenever $|x - c| > R$. [guided] This is a proof by contradiction. Suppose $|x - c| > R$ yet the series $\sum_{n=0}^{\infty} a_n(x - c)^n$ converges. The Boundedness Lemma would then force absolute convergence at every point $y$ with $|y - c| < |x - c|$. Since $R < |x - c|$, we can choose $r'$ with \begin{align*} R < r' < |x - c|. \end{align*} Setting $y = c + r'$, we have $|y - c| = r' < |x - c|$, so the Boundedness Lemma gives absolute convergence at $y$. This means $\sum_{n=0}^{\infty} |a_n| (r')^n$ converges, so $r' \in S$. But $r' > R = \sup S$, which contradicts the fact that no element of $S$ can exceed its supremum. The assumption of convergence at $x$ is therefore false, and the series diverges whenever $|x - c| > R$. Note that this argument tells us nothing about what happens at $|x - c| = R$. The boundary point $R$ may or may not belong to $S$; both situations occur in practice (e.g., $\sum x^n$ diverges at $|x| = 1$, while $\sum x^n/n^2$ converges at $|x| = 1$). [/guided] [/step] [step:Classify the three cases to complete the trichotomy] The preceding two steps show that $\sum_{n=0}^{\infty} a_n(x-c)^n$ converges absolutely for $|x-c| < R$ and diverges for $|x-c| > R$. The three cases in the theorem correspond to the three possible values of $R$: **Case (i): $R = 0$.** The condition $|x - c| < R = 0$ is satisfied by no $x$, so the convergence region contains only $x = c$ (where every power series converges to $a_0$). **Case (ii): $R = \infty$.** The condition $|x - c| < R = \infty$ is satisfied by every $x \in \mathbb{R}$, so the series converges absolutely for all $x$. **Case (iii): $0 < R < \infty$.** The series converges absolutely on the open interval $(c - R, c + R)$ and diverges outside the closed interval $[c - R, c + R]$. Since $R = \sup S$ and $S \subset [0, \infty]$, the value $R$ is uniquely determined. The three cases are mutually exclusive and exhaustive, completing the proof. [guided] The three cases correspond to the three possible values of $R = \sup S \in [0, \infty]$: **Case (i):** $R = 0$. Then $|x - c| < R = 0$ is satisfied by no $x$, so the convergence region contains only $x = c$ (where every power series converges to $a_0$). This occurs, for example, when $a_n = n!$ (the coefficients grow too fast for any nonzero radius). **Case (ii):** $R = \infty$. Then $|x - c| < R = \infty$ holds for every $x \in \mathbb{R}$, so the series converges absolutely for all $x$. This occurs, for example, when $a_n = 1/n!$ (the series for $e^{x-c}$). **Case (iii):** $0 < R < \infty$. The series converges absolutely on the open interval $(c - R, c + R)$ and diverges outside $[c - R, c + R]$. The uniqueness of $R$ follows from the uniqueness of the supremum: given the coefficients $(a_n)_{n=0}^{\infty}$ and center $c$, the set $S$ is uniquely determined, and every subset of $[0, \infty]$ has a unique supremum. The three cases are exhaustive because every element of $[0, \infty]$ falls into exactly one of $\{0\}$, $(0, \infty)$, $\{\infty\}$. The [Cauchy-Hadamard Formula](/theorems/203) provides an explicit computation: $1/R = \limsup_{n \to \infty} |a_n|^{1/n}$. [/guided] [/step]