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.


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

  1. Pingback: Weierstrass M Test and Lebesgue’s Dominated Convergence Theorem | Singapore Maths Tuition

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