> [!NOTE] > Let $\sum_{n=0}^\infty a_{n} z_{n}$ be a power series of $\mathbb{C}$ with a [[Radius of Convergence of Complex Power Series|radius of convergence]] $R>0$. Then for every $r< R$, the sequence of functions $f_{k}:= \sum_{n=0}^{k} a_{n}z^{n}$[[Uniform Convergence of Sequence of Real Functions|converges uniformly]] (converges with respect to the supremum/uniform norm) on $(-r, r)$. ###### Proof We show the result by proving that $\left(f_k\right)$ is uniformly Cauchy in $|z| \leqslant r$. We have (assuming that $j \leqslant k)$ $ \left|f_k(z)-f_j(z)\right|=\left|\sum_{n=j+1}^k a_n z^n\right| \leqslant \sum_{n=j+1}^k\left|a_n\right| r^n \leqslant \sum_{n=j+1}^{\infty}\left|a_n\right| r^n $ Since by assumption $\sum_{n=0}^{\infty}\left|a_n\right| r^n$ is finite, given any $\varepsilon>0$ we can choose $N$ large enough to make $\left|f_k(z)-f_j(z)\right|<\varepsilon$ for all $j, k>N$, concluding the proof. (This proof is essentially an application of the [[Comparison Test for Uniform Convergence of Series of Real Functions (Weierstrass M-test)|Weierstrass M-test]] that we covered a few weeks ago.)