Weierstrass M-test
Let \((f_n)\) be a sequence of functions defined on a set \(E\). Suppose there is a sequence of nonnegative constants \((M_n)\) such that \[ |f_n(x)|\leq M_n \qquad \text{for every }x\in E \]and \( \sum_{n=1}^{\infty}M_n<\infty. \) Then the function series \[ \sum_{n=1}^{\infty}f_n(x) \] converges absolutely and uniformly on \(E\).

First note that saying \(\sum_{n=1}^{\infty}f_n(x)\) converges means that we look at the sequence of partial sums

$$ \begin{align*} S_1(x)&=f_1(x),\\ S_2(x)&=f_1(x)+f_2(x),\\ S_3(x)&=f_1(x)+f_2(x)+f_3(x),\\ &\ \vdots\\ S_N(x)&=\sum_{n=1}^{N}f_n(x). \end{align*} $$

The question becomes whether the sequence of function \(S_1,S_2,S_3,\ldots\) converges to some function \(f\). For a fixed \(x\), the values form an ordinary sequence of numbers:

$$ \begin{align*} S_1(x),S_2(x),S_3(x),\ldots \end{align*} $$

If this numerical sequence has a limit, then

$$ \begin{align*} \sum_{n=1}^{\infty}f_n(x) := \lim_{N\to\infty}S_N(x). \end{align*} $$

For example, at a fixed \(x=x_0\),

$$ \begin{align*} \sum_{n=1}^{\infty}f_n(x_0) = \lim_{N\to\infty} \left[f_1(x_0)+f_2(x_0)+\cdots+f_N(x_0)\right]. \end{align*} $$

while in absolute convergence we examine the series formed from the absolute values:

$$ \begin{align*} |f_1(x)|+|f_2(x)|+|f_3(x)|+\cdots. \end{align*} $$

For a fixed \(x\), define the absolute partial sums

$$ \begin{align*} A_N(x)=\sum_{n=1}^{N}|f_n(x)|. \end{align*} $$

We say \(\sum_{n=1}^{\infty}f_n(x)\) converges absolutely at \(x\) if the numerical sequence

$$ \begin{align*} A_1(x),A_2(x),A_3(x),\ldots \end{align*} $$

converges to a finite number. Equivalently,

$$ \begin{align*} \sum_{n=1}^{\infty}|f_n(x)|=A(x)<\infty \end{align*} $$

Because every term \(|f_n(x)|\) is nonnegative, \(A_N(x)\) is increasing. Therefore, it converges precisely when it is bounded above. (Note that the bound must not depend on \(x\), unless this is pointwise convergence)

Finally, The series converges uniformly to \(f\) on \(E\) if

$$ \begin{align*} \forall\varepsilon>0,\ \exists N \quad\text{such that}\quad n\ge N \Longrightarrow |S_n(x)-f(x)|<\varepsilon \end{align*} $$

for every \(x\in E\).

Now, If we have constants \(M_n\ge 0\) such that \(|f_n(x)|\le M_n\) for every \(x\) in \(E\) and

$$ \begin{align*} \sum_{n=1}^{\infty}M_n<\infty, \end{align*} $$

then the Weierstrass \(M\)-test gives us two conclusions:

  1. For every fixed \(x\in E\), \[ \sum_{n=1}^{\infty}|f_n(x)| \le \sum_{n=1}^{\infty}M_n <\infty, \]so the series converges absolutely at every \(x\).
  2. Because the same constants \(M_n\) work for every \(x\), the series \( \sum_{n=1}^{\infty}f_n(x) \) converges uniformly on \(E\).

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi