In Baby Rudin's PMA, he stated and proved Theorem 8.1 as follows.
In fact, he applied Theorems for sequences to series so that the proof seems a bit hard for readers like me.
So, I have tried a proof using sequences of partial sum as follows.
Proof
It is given that the series (3) converges for $\left|x\right|<R$ and the limit function $f$ is defined by (4).
Thus, it follows that the sequence of partial sum $\{s_n\}, s_n=\sum_{i=0}^{n}{c_i\ x^i}$, converges for $\left|x\right|<R$ and that $\lim_{n\rightarrow\infty}{s_n}=f$.
Since the sum of a finite number of differentiable functions is differentiable, we know that $\{s_n\}$ is differentiable and that $s_n^\prime=\sum_{i=0}^{n}{{ic}_i\ x^{i-1}}$ so that $\lim_{n\rightarrow\infty}{s_n^\prime}=\sum_{i=0}^{\infty}{{ic}_i\ x^{i-1}}$ which is the same as the power series (5).
Let $\varepsilon>0$ be given. For $\left|x\right|\le R-\varepsilon$, we have $$\left|c_nx^n\right|\le\left|c_n\left(R-\varepsilon\right)^n\right|;$$ and since $$\sum{c_n\left(R-\varepsilon\right)^n}$$ converges absolutely (every power series converges absolutely in the interior of its interval of convergence, by the root test), Theorem 7.10 shows the uniform convergence of $\{s_n\}$ on $[-R+\varepsilon,\ R-\varepsilon]$. More ever, since $\{s_n\}$ is continuous on $[-R+\varepsilon,\ R-\varepsilon]$, by Theorem 7.12, so is the limit function $f$ given by (4).
Since $\sqrt[n]{n}\rightarrow1$ as $n\rightarrow\infty$ i.e., ${\limsup}_{n\rightarrow\infty}{\sqrt[n]{n\ }}=1$, we have $${\limsup}_{n\rightarrow\infty}{\sqrt[n]{n\ |c_n|}}={\lim sup}_{n\rightarrow\infty}{\sqrt[n]{n\ }}\ {\limsup}_{n\rightarrow\infty}{\sqrt[n]{|c_n|}}={\limsup}_{n\rightarrow\infty}{\sqrt[n]{|c_n|}},$$ so that the sequences $\{s_n\}$ and $\{s_n^\prime\}$ have the same interval of convergence namely $(-R,R)$.
As in the preceding argument, we can show that, for every $\varepsilon>0$, the sequence $\{s_n^\prime\}$ converges uniformly on $[-R+\varepsilon,\ R-\varepsilon]$.
Thus we have $\{s_n\}$ is differentiable on $[-R+\varepsilon,\ R-\varepsilon], \{s_n(R-\varepsilon)\}$ converges uniformly at some point $R-\varepsilon$ on $[-R+\varepsilon,\ R-\varepsilon]$ and $\{s_n^\prime\}$ converges uniformly on $[-R+\varepsilon,\ R-\varepsilon]$. Then, we can apply Theorem 7.17 to get $$f^\prime\left(x\right)=\lim_{n\rightarrow\infty}{s_n^\prime(x)}=\sum_{i=1}^{\infty}{{ic}_i\ x^{i-1}} $$ on $[-R+\varepsilon,\ R-\varepsilon]$.
It follows that (5) holds if $\left|x\right|\le R-\varepsilon$.
But, given any $x$ such that $\left|x\right|<R$, we can find an $\varepsilon>0$ such that $\left|x\right|<R-\varepsilon$. This shows that (5) holds for $\left|x\right|<R$. Continuity of $f$ follows from Theorem 7.12 or from the existence of $f^\prime$ (Theorem 5.2).\qed
I wonder if my attempt is valid or not. Please help me check. Thanks.