# What's significant in Abel's Theorem Proof in Baby Rudin?

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.

• Why is it uniform on (-1,1)? I think it is only uniform on closed subsets of (-1,1) Feb 5, 2018 at 10:23
• Yes. But [-1+$\epsilon$,1-$\epsilon$] for any $\epsilon>0$ is just (-1,1). Feb 5, 2018 at 10:27
• $x^n$ converges locally uniformly to 0 on (0,1) but not uniformly Feb 5, 2018 at 10:28
• @CalvinKhor Excuse me, what does this imply? I have no exposure to the term ''locally uniform''. Feb 5, 2018 at 10:34
• locally uniform is what i said above: uniform on closed and bounded subsets Feb 5, 2018 at 10:35

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.
• Sorry if I miss something. Isn't "the $\sum_{n=0}^{\infty} c_n$ exists" in the hypothesis of the theorem? Feb 5, 2018 at 11:09
• So basically, if $\sum^{\infty}_{n=0} c_n$ exists, apply M-test and my statement can be proven true. Yes? Feb 5, 2018 at 12:56
• @Math No. The $c_n$ don't need to be positive. Feb 5, 2018 at 13:07
• @Math No. $f (x)$ is absolutely convergent in the interior of its interval of convergence. This does not make the $c_n$ be positive, so the M-test may not apply. But yes, if you know to begin with that $\sum c_n$ converges absolutely, then you can proceed as you suggest. But this case is significantly easier than the general case, where $\sum c_n$ only converges conditionally. Feb 5, 2018 at 13:58