[proofplan] The [radius of convergence](/theorems/273) is defined via the Cauchy--Hadamard formula $1/R = \limsup_{n \to \infty} |c_n|^{1/n}$. The root test establishes absolute convergence when $|x - a| < R$ (by comparison with a geometric [series](/page/Series)) and divergence when $|x - a| > R$ (since the general term fails to tend to zero). [Uniform convergence](/page/Uniform%20Convergence) on compact sub-intervals $[a - r, a + r]$ with $r < R$ follows from the [Weierstrass M-Test](/theorems/261) with $M_n = |c_n| r^n$. [/proofplan] [step:Define the radius of convergence via the Cauchy--Hadamard formula] Define the radius of convergence by \begin{align*} \frac{1}{R} &:= \limsup_{n \to \infty} |c_n|^{1/n}, \end{align*} with the conventions $1/0 = \infty$ and $1/\infty = 0$. [/step] [step:Prove absolute convergence for $|x - a| < R$ by comparison with a geometric series] Suppose $|x - a| < R$. Choose $\rho$ with $|x - a| < \rho < R$, so that $1/\rho > 1/R = \limsup |c_n|^{1/n}$. By the definition of limsup, there exists $N \in \mathbb{N}$ such that $|c_n|^{1/n} < 1/\rho$ for all $n \geq N$. Equivalently, $|c_n| < \rho^{-n}$ for $n \geq N$. Therefore: \begin{align*} |c_n (x - a)^n| &= |c_n| \cdot |x - a|^n < \rho^{-n} \cdot |x - a|^n = \left(\frac{|x - a|}{\rho}\right)^n. \end{align*} Since $|x - a|/\rho < 1$, the geometric [series](/page/Series) $\sum_{n=0}^\infty (|x - a|/\rho)^n$ converges. By the comparison test, $\sum_{n=0}^\infty |c_n (x - a)^n|$ converges, establishing absolute convergence. [/step] [step:Prove divergence for $|x - a| > R$ via the divergence test] Suppose $|x - a| > R$. Then $1/|x - a| < 1/R = \limsup |c_n|^{1/n}$, so there exist infinitely many indices $n$ with $|c_n|^{1/n} > 1/|x - a|$, i.e., $|c_n| > |x - a|^{-n}$. For these indices: \begin{align*} |c_n (x - a)^n| &= |c_n| \cdot |x - a|^n > 1. \end{align*} Since the general term does not converge to zero, the [series](/page/Series) diverges by the divergence test. [/step] [step:Establish uniform convergence on $[a - r, a + r]$ via the Weierstrass $M$-test] Fix $0 < r < R$. For each $n \in \mathbb{N}$ and each $x \in [a - r, a + r]$: \begin{align*} |c_n (x - a)^n| &\leq |c_n| r^n =: M_n. \end{align*} Since $r < R$, the point $x = a + r$ satisfies $|x - a| = r < R$, so $\sum_{n=0}^\infty |c_n| r^n$ converges by the absolute convergence established above. The [Weierstrass M-Test](/theorems/261) with the bounds $M_n = |c_n| r^n$ gives absolute and [uniform convergence](/page/Uniform%20Convergence) of $\sum_{n=0}^\infty c_n (x - a)^n$ on $[a - r, a + r]$. [/step] [step:Verify uniqueness of $R$] The value $R$ is uniquely determined: if $R' < R$, there exists a point $x$ with $R' < |x - a| < R$ where the [series](/page/Series) converges (by absolute convergence inside $R$), contradicting divergence outside $R'$. Similarly, $R' > R$ leads to a point where the series both diverges and converges. Therefore $R' = R$. [/step]