Weierstrass M-test Proof and Special Case of Abel’s Theorem

First, let us recap what is Weierstrass M-test:

Weierstrass M-test:

Let $\{f_n\}$ be a sequence of real (or complex)-valued functions defined on a set A, and let $\{M_n\}$ be a sequence satisfying $\forall n\in\mathbb{N}, \forall x\in A$

$|f_n (x)|\leq M_n$ , and also $\sum_{n=1}^\infty M_n=M<\infty$ .

Then, $\sum_{n=1}^\infty f_n(x)$ converges uniformly on A (to a function f).

Proof:

Let $\epsilon >0$ . $\exists N\in\mathbb{N}$ such that $m\geq N$ implies $|M-\sum_{n=1}^m M_n|<\epsilon$ .

For $m\geq N, \forall x\in A$ ,

$\begin{aligned} |f(x)-\sum_{n=1}^m f_n(x)|&=|\sum_{n=m+1}^\infty f_n (x)|\\ &\leq\sum_{n=m+1}^\infty |f_n (x)|\\ &\leq \sum_{n=m+1}^\infty M_n\\ &=|M-\sum_{n=1}^m M_n|\\ &<\epsilon \end{aligned}$

Thus, $\sum_{n=1}^\infty f_n (x)$ converges uniformly.

Application to prove Abel’s Theorem (Special Case):

Consider the special case of Abel’s Theorem where all the coefficients $a_i$ are of the same sign (e.g. all positive or all negative).

Then, for $x\in [0,1]$ ,

$|a_n x^n|\leq |a_n|:=M_n$

Then by Weierstrass M-test, $\sum_{n=1}^\infty a_n x^n$ converges uniformly on [0,1] and thus $\lim_{x\to 1^-} \sum_{n=1}^\infty a_n x^n=\sum_{n=1}^\infty a_n$ .