[proofplan] We show that on every compact subinterval $[c-r, c+r]$ with $r < R$, the [power series](/page/Power%20Series) converges uniformly by applying the [Weierstrass M-Test](/theorems/272). Since each partial sum is a polynomial — hence continuous — the [Uniform Limit Theorem](/theorems/258) yields continuity of $f$ on each such subinterval. Because every point $x_0 \in (c - R, c + R)$ belongs to some $[c-r, c+r]$, continuity extends to the full interval. [/proofplan] [step:Bound each summand uniformly on $[c - r, c + r]$ for $r < R$] Fix $x_0 \in (c - R, c + R)$ and choose $r$ with $|x_0 - c| < r < R$. For each $n \geq 0$, define \begin{align*} g_n: [c - r, c + r] &\to \mathbb{R} \\ x &\mapsto a_n(x - c)^n. \end{align*} For every $x \in [c - r, c + r]$ and every $n \geq 0$, \begin{align*} |g_n(x)| = |a_n| \cdot |x - c|^n \leq |a_n| \cdot r^n, \end{align*} where the inequality holds because $|x - c| \leq r$ on the compact interval $[c - r, c + r]$. Set $M_n := |a_n| \cdot r^n$, so that $\sup_{x \in [c-r, c+r]} |g_n(x)| \leq M_n$. [guided] The goal of this step is to prepare the hypotheses required by the [Weierstrass M-Test](/theorems/272). That test demands a dominating numerical series $\sum M_n$ that bounds each function $g_n$ uniformly over the domain. To construct such bounds, we exploit the fact that on any *closed* subinterval $[c - r, c + r]$ with $r < R$, the factor $|x - c|^n$ never exceeds $r^n$. Fix an arbitrary point $x_0 \in (c - R, c + R)$. Since $(c - R, c + R)$ is open, there exists $r$ satisfying $|x_0 - c| < r < R$, so that $x_0 \in [c - r, c + r] \subset (c - R, c + R)$. For each $n \geq 0$, define the $n$-th summand as a function on this compact interval: \begin{align*} g_n: [c - r, c + r] &\to \mathbb{R} \\ x &\mapsto a_n(x - c)^n. \end{align*} For every $x \in [c - r, c + r]$, we have $|x - c| \leq r$, so \begin{align*} |g_n(x)| = |a_n| \cdot |x - c|^n \leq |a_n| \cdot r^n. \end{align*} Setting $M_n := |a_n| \cdot r^n$ gives the uniform bound $\sup_{x \in [c-r, c+r]} |g_n(x)| \leq M_n$ that the [Weierstrass M-Test](/theorems/261) requires. [/guided] [/step] [step:Verify that $\sum M_n$ converges and apply the Weierstrass M-Test] Since $0 < r < R$ and $R$ is the [radius of convergence](/theorems/273), the series $\sum_{n=0}^{\infty} |a_n| \cdot r^n$ converges — this follows directly from the definition of [radius of convergence](/theorems/265), which guarantees absolute convergence at every point strictly inside the radius. Hence $\sum_{n=0}^{\infty} M_n < \infty$. We now verify the hypotheses of the [Weierstrass M-Test](/theorems/272): the functions $g_n: [c - r, c + r] \to \mathbb{R}$ satisfy $\sup_{x \in [c - r, c + r]} |g_n(x)| \leq M_n$ for all $n \geq 0$, and $\sum_{n=0}^{\infty} M_n$ converges. By the [Weierstrass M-Test](/theorems/272), the series $\sum_{n=0}^{\infty} g_n = \sum_{n=0}^{\infty} a_n(x - c)^n$ converges uniformly on $[c - r, c + r]$. [guided] It remains to verify that the dominating series $\sum_{n=0}^{\infty} M_n = \sum_{n=0}^{\infty} |a_n| \cdot r^n$ actually converges. The key is that $r < R$. By the [definition of radius of convergence](/theorems/273), the power series $\sum_{n=0}^{\infty} a_n (x - c)^n$ converges absolutely whenever $|x - c| < R$. In particular, evaluating at $x = c + r$ (so that $|x - c| = r < R$) yields \begin{align*} \sum_{n=0}^{\infty} |a_n (c + r - c)^n| = \sum_{n=0}^{\infty} |a_n| \cdot r^n < \infty. \end{align*} This is precisely $\sum_{n=0}^{\infty} M_n < \infty$. We now check the two hypotheses of the [Weierstrass M-Test](/theorems/272): 1. **Uniform bound:** For every $n \geq 0$ and every $x \in [c - r, c + r]$, we have $|g_n(x)| \leq M_n$ (established in the previous step). 2. **Convergent majorant:** $\sum_{n=0}^{\infty} M_n < \infty$ (established above). Both conditions are satisfied. By the [Weierstrass M-Test](/theorems/272), the series $\sum_{n=0}^{\infty} g_n(x) = \sum_{n=0}^{\infty} a_n(x - c)^n$ converges uniformly on $[c - r, c + r]$. Why is [uniform convergence](/page/Uniform%20Convergence) — rather than mere pointwise convergence — essential here? Because continuity is a *local* property involving limits, and the passage from "each partial sum is continuous" to "the limit is continuous" requires the convergence to be *uniform*. Pointwise convergence alone does not preserve continuity: for instance, the pointwise limit of $x^n$ on $[0,1]$ is discontinuous at $x = 1$. [/guided] [/step] [step:Observe that each partial sum is a polynomial, hence continuous] For each $N \geq 0$, define the $N$-th partial sum: \begin{align*} S_N: [c - r, c + r] &\to \mathbb{R} \\ x &\mapsto \sum_{n=0}^{N} a_n (x - c)^n. \end{align*} The function $S_N$ is a polynomial of degree at most $N$ in the variable $(x - c)$, and therefore in $x$. Since every polynomial function from $\mathbb{R}$ to $\mathbb{R}$ is continuous (it is a finite sum of products of continuous functions), $S_N$ is continuous on $[c - r, c + r]$. [/step] [step:Conclude continuity on $[c - r, c + r]$ via the Uniform Limit Theorem] We have established: (i) $S_N \to f$ uniformly on $[c - r, c + r]$, and (ii) each $S_N$ is continuous on $[c - r, c + r]$. These are precisely the hypotheses of the [Uniform Limit Theorem](/theorems/258), which asserts that the uniform limit of a sequence of continuous functions on a [metric space](/page/Metric%20Space) is continuous. Since $[c - r, c + r] \subset \mathbb{R}$ is a metric space under the standard metric, the theorem applies and yields that $f$ is continuous on $[c - r, c + r]$. [guided] We apply the [Uniform Limit Theorem](/theorems/258). That theorem requires: 1. **A metric space:** The domain $[c - r, c + r]$ is a subset of $\mathbb{R}$ equipped with the standard Euclidean metric $d(x, y) = |x - y|$, hence a metric space. 2. **A sequence of continuous functions:** Each partial sum $S_N: [c - r, c + r] \to \mathbb{R}$ is a polynomial, hence continuous on $[c - r, c + r]$. 3. **Uniform convergence:** The [Weierstrass M-Test](/theorems/264) established that $S_N \to f$ uniformly on $[c - r, c + r]$. All three conditions hold. The [Uniform Limit Theorem](/theorems/258) therefore guarantees that $f$ is continuous on $[c - r, c + r]$. To make the mechanism fully explicit: fix $\varepsilon > 0$. By uniform convergence, there exists $N \in \mathbb{N}$ such that \begin{align*} |f(x) - S_N(x)| < \frac{\varepsilon}{3} \quad \text{for all } x \in [c - r, c + r]. \end{align*} Since $S_N$ is continuous at $x_0$, there exists $\delta > 0$ such that \begin{align*} |S_N(x) - S_N(x_0)| < \frac{\varepsilon}{3} \quad \text{whenever } |x - x_0| < \delta \text{ and } x \in [c - r, c + r]. \end{align*} For any such $x$, the triangle inequality gives \begin{align*} |f(x) - f(x_0)| &\leq |f(x) - S_N(x)| + |S_N(x) - S_N(x_0)| + |S_N(x_0) - f(x_0)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align*} The first and third terms are controlled by the *uniform* bound on $|f - S_N|$ (note: both $x$ and $x_0$ lie in $[c - r, c + r]$, so the same $N$ works for both points). The middle term is controlled by the continuity of the polynomial $S_N$ at $x_0$. This is the standard "$\varepsilon/3$ argument" that underlies the Uniform Limit Theorem. [/guided] [/step] [step:Extend continuity to all of $(c - R, c + R)$] Since $x_0 \in (c - R, c + R)$ was arbitrary, and for each such $x_0$ we found $r$ with $|x_0 - c| < r < R$ on which $f$ is continuous, the function $f$ is continuous at every point of $(c - R, c + R)$. Hence $f$ is continuous on $(c - R, c + R)$. [/step]