A stuck in Theorem 8.2 of Baby Rudin's PMA The following is Rudin's Theorem $8.2$ and its proof.

The proof seems a bit hard for readers like me. So, I have tried an approach to its proof using partial sums and applying Theorem $7.8$ & Theorem $7.12$ as follows.
Proof
Consider a sequence $\left\{s_n\right\}$ of partial sums $s_n(x)=\sum_{i=0}^{n}{c_i\ x^i}$, where $\left|x\right|<1$ and where $\sum {c_n}$ converges.
Since  $\sum\ c_n$ converges, for every $\varepsilon>0$, there exists an integer $N$ such that $\sum_{i=m}^{n}\left|{c_i}\right|<\varepsilon$ if $m,n>N$.
Then,
$\left|s_n\left(x\right)-s_m\left(x\right)\right|=\left|\sum_{i=0}^{n}{c_ix^i}-\sum_{i=0}^{m}{c_ix^i}\right|$
$$= \left|\sum_{i=m+1}^{n}c_i x^i\right|\le\sum_{i=m+1}^{n}\left|c_i\right|\le\sum_{i=m}^{n}\left|c_i\right|<\varepsilon,\  \text{if}\ m,n>N.$$
Thus, by Theorem $7.8$, Cauchy-criterion, ${s_n}$ converges uniformly to $f$ given by $f(x)=\sum_{n=0}^{\infty}{c_n\ x^n\ }$.
Since $s_n$ is the sum of a finite number of continuous functions on $[-1,1]$, $s_n$ is continuous on $[-1,1]$, too.
Then, by Theorem $7.12$, the limit function $f$ is continuous on $[-1,1]$.
Then, we obtain
$$\lim_{x→1}\;f(x)=\lim_{x→1}⁡∑_{n=0}^∞\;c_n x^n =∑_{n=0}^∞c_n .$$
\qed
I am wondering if my attempt is valid or not and is it isn't please suggest me. Thanks.
The following are the theorems I used in my attempt.

 A: Your proof has two issues:

*

*In your application of 7.10, you've assumed $\sum c_n$ converges absolutely.

*In your final steps, you've only written that $f$ is continuous on $(-1,1)$, which in no way justifies the evaluation of the limit as $x \to 1$.

Of these, the former is devastating and the latter is mostly cosmetic.
That is, dealing with them in reverse:

*

*Your applications of 7.10 and 7.12 could have just used $E = [-1,1]$ instead of $E = (-1,1)$.  With this change, you would be able to conclude from $f$ being continuous on $[-1,1]$ that the limit $\lim_{x\to 1^-} f(x) = f(1)$, which would be exactly what you wanted to show, if not for the fact...

*Theorem 8.2 only assumes $\sum c_n$ converges, not that it converges absolutely.  With this, your initial application of 7.10 is invalid.

The good news is that your proof can be immediately modified to prove the corollary of 8.2 which is identical to 8.2 except that we strengthen the assumption to $\sum c_n$ converges absolutely.  The bad news is that this weakening of 8.2 is much less useful, and obviously it fails to provide an alternative proof of 8.2 itself.
A: The following is the proof of Theorem 8.2 in which some modifications are added to the original one in order for the reader like me to see.
Proof:
Let $ s_n=c_0+\dots+c_n,\ \ s_(-1)=0.$ Then
$$\sum_{n=0}^{m}{c_nx^n}=\sum_{n=0}^{m}{\left(s_n-s_{n-1}\right)x^n}=\sum_{n=0}^{m}{s_nx^n}-x\sum_{n=0}^{m}{s_{n-1}x}^{n-1}=\left(1-x\right)\sum_{n=0}^{m-1}{s_nx^n}+s_mx^m-s_{-1}=\left(1-x\right)\sum_{n=0}^{m-1}{s_nx^n}+s_mx^m.$$
For $\left|x\right|<1$, we let $m\rightarrow\infty$ and obtain
(9) $f\left(x\right)=\lim_{m\rightarrow\infty}{\sum_{n=0}^{m}{c_nx^n}}=\lim_{m\rightarrow\infty}{\left(1-x\right)\sum_{n=0}^{m-1}{s_nx^n}}=\left(1-x\right)\sum_{n=0}^{\infty}{s_nx^n}.$
Since $\lim_{n\rightarrow\infty}{s_n}=\sum\ c_n$ converges, suppose $s=\lim_{n\rightarrow\infty}{s_n}$. Let $\varepsilon>0$ be given. Choose $N$ so that $n>N$ implies
$$\left|s-s_n\right|<\frac{\varepsilon}{2}.$$
Then, since
$$\left(1-x\right)\sum_{n=0}^{\infty}x^n=\left(1-x\right)\times\frac{1}{1-x}=1,$$
we obtain from (9)
$$
\left|f\left(x\right)-s\right|=\left|\left(1-x\right)\sum_{n=0}^{\infty}{s_nx^n}-\left(1-x\right)\sum_{n=0}^{\infty}x^ns\right|=\left|\left(1-x\right)\sum_{n=0}^{\infty}{(s_n-s)x^n}\right|\le\left|\left(1-x\right)\sum_{n=0}^{N}{\left(s_n-s\right)x^n}\right|+\left|\left(1-x\right)\sum_{n=N+1}^{\infty}{\left(s_n-s\right)x^n}\right|<\left|\left(1-x\right)\sum_{n=0}^{N}{\left(s_n-s\right)x^n}\right|+\frac{\varepsilon}{2}\ \left|\left(1-x\right)\sum_{n=N+1}^{N}x^n\right|\le\left|\left(1-x\right)\sum_{n=0}^{N}{\left(s_n-s\right)x^n}\right|+\frac{\varepsilon}{2}.
$$
If $x>1-\delta$, for some suitably so chosen $\delta$ that $\delta\ \left|\sum_{n=0}^{N}{\left(s_n-s\right)x^n}\right|\le\frac{\varepsilon}{2}$, we get
$$\left|f\left(x\right)-s\right|\le\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$
Note that $x>1-\delta$ and $|x|<1$ means $|1-x|<\delta$.
Thus, it follows that for every $\varepsilon>0$ there exists $\delta>0$ such that $\left|1-x\right|<\delta\Rightarrow\left|f\left(x\right)-s\right|\le\varepsilon.$
Thus, we conclude that $\lim_{x\rightarrow1}{f(x)}=s=\sum_{n=0}^{\infty}c_n$.
A: In the following proof I have used Theorem 7.11 instead of Theorem 7.12.
Proof
Consider a sequence $\left\{s_n\right\}$ of partial sums $s_n(x)=\sum_{i=0}^{n}{c_i\ x^i}$, where $\left|x\right|<1$ and where $\sum {c_n}$ converges.
Since  $\sum\ c_n$ converges, for every $\varepsilon>0$, there exists an integer $N$ such that $\sum_{i=m}^{n}\left|{c_i}\right|<\varepsilon$ if $m,n>N$.
Then,
$\left|s_n\left(x\right)-s_m\left(x\right)\right|=\left|\sum_{i=0}^{n}{c_ix^i}-\sum_{i=0}^{m}{c_ix^i}\right|$
$$= \left|\sum_{i=m+1}^{n}c_i x^i\right|\le\sum_{i=m+1}^{n}\left|c_i\right|\le\sum_{i=m}^{n}\left|c_i\right|<\varepsilon,\  \text{if}\ m,n>N.$$
Thus, by Theorem $7.8$, Cauchy-criterion, ${s_n}$ converges uniformly to $f$ given by $f(x)=\sum_{n=0}^{\infty}{c_n\ x^n\ }$.
Since $1$ is a limit point of $(-1,1)$ and $\lim_{x→1}\;⁡s_n (x)=∑_{i=0}^n\;c_i =A_n$, then, by theorem $7.11$, ${A_n}$ converges and
$$\lim_{x→1}⁡f(x)=\lim_{n→∞}\;A_n=\sum_{n=0}^∞\;c_n .$$
\qed
The following is Theorem 7.11

