> [!NOTE] **Theorem (Continuity of power series)** > Let $\sum_{0}^{\infty}a_{n} x^{n}$ be a [[Power Series|univariate real power series about zero]] with [[Radius of Convergence of Complex Power Series|radius of convergence]] $R$. Then the [[Real Function|real function]] $x \mapsto \sum_{0}^{\infty}a_{n} x^{n}$is [[Continuous Real Function|continuous]] on the interval $(-R,R)$. **Proof**: Let $c \in (-R,R)$. WTS for all $\varepsilon>0,$ there exists $\delta>0$ such that for all $x\in (-R,R),$ $|x-c|<\delta \implies \left\lvert \sum a_{n}x^{n} - \sum a_{n}c^{n} \right\rvert < \varepsilon. $ Let $|c|<T<R$. Then by [[Radius of Convergence of Absolute Real Power Series About Zero]], the series $\sum |a_{n}|T^{n}$ [[Convergent Real Series|converges]], so for each $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that $ \sum_{N+1}^{\infty} \lvert a_{n}T^{n} \rvert <\varepsilon/3.$Now if $|x-c|< \delta_{1}:= T-|c|$ we have $|x|<T.$ Hence $ \sum_{N+1}^{\infty} \lvert a_{n}x^{n} \rvert < \frac{\varepsilon}{3} \quad \text{ and } \quad \sum_{N+1}^{\infty} \lvert a_{n}c^{n} \rvert < \frac{\varepsilon}{3}.$ The sum $\sum_{0}^{N} a_{n} x^{n}$ is a polynomial in $x$ and polynomials are continuous so there is some $\delta_{2}>0$ such that if $|x-c|<\delta_{2}$, $\left\lvert \sum_{0}^{N } a_{n} x^{n} - \sum_{0}^{N} a_{n} c^{n} \right\rvert < \frac{\varepsilon}{3}.$ Therefore if we choose $|x-c|<\delta:=\min(\delta_{1},\delta_{2})$ we have $\begin{align} \left\lvert \sum_{0}^{\infty } a_{n} x^{n} - \sum_{0}^{\infty} a_{n} c^{n} \right\rvert & \leq \left\lvert \sum_{N+1}^{\infty}a_{n} x_{n } \right\rvert + \left\lvert \sum_{0}^{N } a_{n} x^{n} - \sum_{0}^{N} a_{n} c^{n} \right\rvert + \left\lvert \sum_{N+1}^{\infty}a_{n} c_{n } \right\rvert \\ & \leq \sum_{N+1}^{\infty} |a_{n}| |x|^{n} + \left\lvert \sum_{0}^{N } a_{n} x^{n} - \sum_{0}^{N} a_{n} c^{n} \right\rvert + \sum_{N+1}^{\infty} |a_{n}| |c|^{n} < \varepsilon. \end{align}$