Refering to Theorem 8.2 in Baby Rudin
8.2 Theorem Suppse $\sum c_n$ converges. Put $$f(x)=\sum_{n=0}^{\infty} c_nx^n \ \ \ \ (-1<x<1)$$ Then $$\lim_{x\rightarrow1}f(x)=\sum_{n=0}^{\infty}c_n$$
The proof in Rudin is that outlined in Wikipedia
However, Factoring out $(1-x)$ seems not natural for me.
The theorem looks like a extension from Theorem 8.1.
Indeed, by imitating the proof of Rudin 7.11 $$|f(t)-\sum^{\infty}_{n=0}c_n|\leq|\sum_{n=0}^{\infty}c_n-\sum_{n=0}^{N}c_n|+|\sum_{n=0}^{N}c_n-\sum_{n=0}^{N}c_nt^n|+|\sum_{n=0}^{N}c_nt^n-f(t)|$$ where $t \in(-1,1)$
1.The first term can be arbitrary small for large N since $\sum c_n$ converges.
2.The second term can be small when $t\rightarrow1$ since polynomial is continuous.
3.Since f(x) is uniform convergent on $(-1,1)$, the third term can also be arbitrary small.
Then, I can also conclude the same result as in Theorem 8.2.
Could anyone kindly explain why Rudin uses a different approach (like factoring out (1-x)) or my reasoning has flaws, if there is any ?
Summary of the discussion :
The problem in my proof arises from the third term in the inequality, where I mistake the order to take limits (I implicitly make $t \rightarrow 1$ and then $ N \rightarrow \infty$. This is wrong).
The motivation (I guess) in Rudin's proof is from Rudin Theorem 3.42, where we study the criteria to test conditional convergent series. Theorem 8.2 has a similar situation, the sum $\sum c_n$ may very well be not absolutely convergent.
You claim that $$\left | \sum_{n=0}^N c_n x^n - f(x) \right| \xrightarrow[N\to\infty]{} 0 \quad \text{uniformly in }(-1,1) $$ this is not necessarily true. For it to have any chance to be true, it must make reference to the extra assumption that $\sum_{n=0}^\infty c_n$ exists, because its not true for arbitrary power series that converge (locally uniformly) on $(-1,1)$.
For instance, consider $f(x) = \sum_{n=0}^\infty x^n = \frac1{1-x}$. Its partial sums cannot converge uniformly on any $(1-\epsilon,1)$ since the partial sums are bounded but the limit is unbounded.