[proofplan] Parts (1) and (2) follow from the root test applied to the numerical [series](/page/Series) $\sum |c_n(x-a)^n|$, where the $n$-th root is $|c_n|^{1/n}|x-a|$. Convergence holds when this limsup is strictly less than $1$, i.e., $|x - a| < R = 1/\limsup |c_n|^{1/n}$, and divergence holds when it exceeds $1$. Part (3) follows from the Weierstrass $M$-test (Theorem 264) with $M_n = |c_n| r^n$. [/proofplan] [step:Compute the limsup of the $n$-th root of the general term] Define $\alpha = \limsup_{n \to \infty} |c_n|^{1/n} \in [0, \infty]$ and set $R = 1/\alpha$. For each $x \in \mathbb{R}$, the general term $b_n = c_n(x-a)^n$ satisfies \begin{align*} |b_n|^{1/n} &= |c_n|^{1/n} \cdot |x - a|. \end{align*} Since $|x - a|$ is a non-negative constant, $\limsup_{n \to \infty} |b_n|^{1/n} = \alpha \cdot |x - a|$. [/step] [step:Apply the root test to establish absolute convergence for $|x - a| < R$] If $|x - a| < R = 1/\alpha$, then $\alpha \cdot |x - a| < 1$. By the previous step, $\limsup |b_n|^{1/n} < 1$. The root test for numerical [series](/page/Series) gives absolute convergence of $\sum |b_n| = \sum |c_n| \cdot |x - a|^n$. [/step] [step:Apply the root test to establish divergence for $|x - a| > R$] If $|x - a| > R = 1/\alpha$, then $\alpha \cdot |x - a| > 1$. By the limsup computation, $\limsup |b_n|^{1/n} > 1$, so $|b_n| > 1$ for infinitely many $n$. The terms do not tend to zero, so the [series](/page/Series) diverges. [/step] [step:Establish uniform convergence on $[a - r, a + r]$ via the Weierstrass $M$-test] Fix $0 < r < R$. For $x \in [a - r, a + r]$, $|c_n(x - a)^n| \leq |c_n| r^n =: M_n$. Since $r < R = 1/\alpha$, the root test gives $\limsup |c_n|^{1/n} r = \alpha r < 1$, so $\sum M_n = \sum |c_n| r^n$ converges. The [Weierstrass M-Test](/theorems/264) gives absolute and [uniform convergence](/page/Uniform%20Convergence) on $[a - r, a + r]$. [/step] [step:Confirm uniqueness of $R$] The value $R$ is uniquely determined by the root test characterisation: any $R_1$ satisfying absolute convergence for $|x - a| < R_1$ and divergence for $|x - a| > R_1$ must equal $R$, since otherwise there would exist a point where the [series](/page/Series) both converges and diverges. [/step]