[proofplan] The argument proceeds in two stages. First, we show absolute convergence inside the radius $R = \sup\{|x - a| : \sum c_n(x-a)^n \text{ converges}\}$ by bounding the terms with a geometric [series](/page/Series) when a convergent point at larger radius exists. Second, the Weierstrass $M$-test ([Theorem 272](/theorems/272)) upgrades absolute convergence on closed sub-intervals to [uniform convergence](/page/Uniform%20Convergence). The [root test](/theorems/175) provides the quantitative link to the Cauchy--Hadamard formula. [/proofplan] [step:Define $R$ and prove absolute convergence for $|x - a| < R$] Define $R = \sup\{|x - a| : \sum_{n=0}^\infty c_n(x-a)^n \text{ converges}\} \in [0, \infty]$. Suppose $|x - a| < R$. By definition of the supremum, there exists $y$ with $|y - a| > |x - a|$ such that $\sum c_n(y - a)^n$ converges. Since the [series](/page/Series) converges, the terms $c_n(y-a)^n$ are bounded: there exists $C > 0$ with $|c_n(y-a)^n| \leq C$ for all $n$. Set $\rho = |x-a|/|y-a| < 1$. Then \begin{align*} |c_n(x-a)^n| &= |c_n(y-a)^n| \cdot \left|\frac{x-a}{y-a}\right|^n \leq C \rho^n. \end{align*} Since $\sum C\rho^n$ converges (geometric series with ratio $\rho < 1$), the [comparison test](/theorems/173) gives absolute convergence. If $|x - a| > R$, the series diverges by definition of $R$ as the supremum. [guided] The key idea is that convergence at a point $y$ with $|y - a| > |x - a|$ implies absolute convergence at $x$. If $\sum c_n(y-a)^n$ converges, its terms must be bounded: $|c_n(y-a)^n| \leq C$. Then for $|x - a| < |y - a|$: \begin{align*} |c_n(x-a)^n| &= |c_n(y-a)^n| \cdot \left(\frac{|x-a|}{|y-a|}\right)^n \leq C \rho^n, \end{align*} where $\rho = |x-a|/|y-a| < 1$. The series $\sum C\rho^n$ is a convergent geometric series, so absolute convergence follows by comparison. This argument shows that the set of radii at which the series converges forms an interval, justifying the definition of $R$ as its supremum. [/guided] [/step] [step:Establish uniform convergence on $[a - r, a + r]$ via the Weierstrass $M$-test] Fix $0 < r < R$. On $[a - r, a + r]$, each term satisfies $|c_n(x - a)^n| \leq |c_n| r^n$ for all $x$, so set $M_n = |c_n| r^n$. Since $r < R$, the [series](/page/Series) $\sum |c_n| r^n$ converges absolutely (by the previous step, applied at any point with $|x - a| = r$). The [Weierstrass M-Test](/theorems/272) with bounds $M_n = |c_n| r^n$ gives absolute and [uniform convergence](/page/Uniform%20Convergence) on $[a - r, a + r]$. [/step] [step:Derive the Cauchy--Hadamard formula from the root test] The Cauchy--Hadamard formula gives $1/R = \limsup_{n \to \infty} |c_n|^{1/n}$. This follows from the root test applied to the [series](/page/Series) $\sum |c_n| |x - a|^n$: the $n$-th root of the general term is $|c_n|^{1/n} |x - a|$, and the root test gives convergence when $\limsup_{n \to \infty} |c_n|^{1/n} |x - a| < 1$, i.e., $|x - a| < 1/\limsup |c_n|^{1/n}$, and divergence when the limsup exceeds $1$. Comparing with the convergence/divergence behaviour established above identifies $R = 1/\limsup |c_n|^{1/n}$. [/step]