Power series converge uniformly on their radius of convergence

Write \(S_N(z) \da \sum_{0\leq k\leq N} c_k (z-z_0)^k\) and \(S \da \lim_{N\to\infty} S_N\). Suppose \(R\da \inverseof{\qty{\limsup_k \abs{c_k}^{1\over k} }}\) is the radius of convergence and let \(r\leq R\), we’ll show \(S_N\to S\) uniformly on any disc \(\abs{z-z_0}< r\).

Use the \(M\dash\)test: \(\sum f_k\) converges if \(\norm{f_k}_\infty\leq M_k\) where \(\ts{M_k}\in \ell^1(\NN)\). Define \(f_k \da c_k (z-z_0)^k\), then \[ \norm{f_k}_\infty =\sup_{\abs{z-z_0}\leq r} \abs{c_k(z-z_0)^k} \leq \abs{c_k} r^k \da M_k .\] Then \[ \sum_{k\geq 0} M_k = \sum_{k\geq 0} \abs{c_k} r^k ,\] and the claim is that this converges.

Note that since \(r\leq R\), we have convergence of \[ \sum_{k\geq 0} c_k r^k .\] Recall that root test: \[ \sum_k a_k \text{ converges absolutely if } \limsup_k \abs{a_k}^{1\over k} < 1 .\] Here we take \(a_k \da c_k r^k\), then \[ \limsup_k \abs{a_k}^{1\over k} &\da \limsup_k \abs{c_k r^k}^{1\over k} \\ &= \limsup_k \abs{c_k}^{1\over k} r \\ &< \limsup_k \abs{c_k}^{1\over k} R\\ &\da \limsup_k \abs{c_k}^{1\over k} \inverseof{\qty{\limsup_{k} \abs{c_k}^{1\over k} }} \\ &= 1 ,\] so \(\sum_k \abs{c_k r^k} < \infty\). Thus \(\ts{M_k}\in \ell^1(\NN)\), and so \(\sum_k f_k\) converges uniformly and absolutely on \(\abs{z-z_0} = r < R\).