An introduction to the Weierstrass M-test: Part I

In order to make these posts accessible to first-year undergraduate students, I am going to be working with real-valued functions defined on intervals in $\R$. I'm also going to avoid focusing on uniform convergence for sequences of functions, and instead focus on absolute convergence of series of functions, and how this fits with continuity. I'll then start to look at differentiability.

For those interested in my approach to teaching uniform convergence for sequences of functions, I have a video about this on YouTube (at https://www.youtube.com/watch?v=mB4Yny0T3HA). Students watching this video should be warned that my non-standard notion of "absorption" (intended to help students understand convergence of sequences properly) has not really caught on! Other lecturers are unlikely to know about this notion of absorption, and so you should avoid using that terminology unless you explain it locally. However, if you want to know more about my absorption approach, see my blog post at https://explaining-maths.blogspot.com/2021/09/quantifier-packaging-when-teaching.html

Notation:

For convenience, we denote the set  \(\N \cup \{0\}\) of non-negative integers by \(\N_0.\)

Throughout these posts, \(I\) will be a non-degenerate interval in \(\R\), and \((f_n)_{n=0}^\infty\) will be a sequence of functions from \(I\) to \(\R\). 

(For example, we might have \(f_n(x)=x^n/n!\) for each \(n\in\N_0\) and \(x \in I\).)

It is slightly more common to start from \(n=1\), but if we are going to work with power series, starting from \(n=0\) is probably a slightly better fit.

In this post, we are going to give a sufficient condition for the convergence, for all \(x \in I\), of the series

 \(\Jdisplay\sum_{n=0}^\infty f_n(x)\,.\)

We are also going to make a preliminary investigation, in this setting, of the properties of the function

\(\Jdisplay x \mapsto \sum_{n=0}^\infty f_n(x)\,.\)

We can certainly look at the partial sums of the series. So for each \(m \in \N_0\) we define the function \(S_m: I\to \R\) by \(S_m = \sum_{n=0}^m f_n\,.\) So, for each \(x \in I\), we have

\(\Jdisplay S_m(x) = \sum_{n=0}^m f_n (x)\,.\)

For the rest of this post, let \((M_n)_{n=0}^\infty\) be a sequence of non-negative real numbers such that

 \(\Jdisplay\sum_{n=0}^\infty M_n < \infty\,,\)

i.e., this series (with non-negative real terms) is convergent.

For each \(m \in \N_0\), set \(R_m=\sum_{n=m+1}^\infty M_n\,.\)

Then \(R_m \to 0\) as \(m \to \infty,\) because the "tails of a convergent series" always tend to \(0.\)

The following theorem is part of the result known as the Weierstrass M-test 

Theorem 1 (Part of the Weierstrass M-test)

Suppose that, for all \(n \in \N_0\) and all \(x \in I\), we have

\(\Jdisplay |f_n(x)| \leq M_n\,.\)

Then, for all \(x \in I\), the series

 \(\Jdisplay\sum_{n=0}^\infty f_n(x)\)

is absolutely convergent, and so we may define \(f:I \to \R\) by

 \(\Jdisplay f(x)= \sum_{n=0}^\infty f_n(x)\,.\)

Then for all \(m \in \N_0\) and all \(x \in I\) we have

\(\Jdisplay |S_m(x) - f(x)| \leq R_m\,\)

(where \(R_m\) is as defined above). 

Comments

Note that the real numbers \(R_m\) do not depend on \(x\).

The sequence \((R_m)_{m=0}^\infty\) is non-increasing (\(R_0\geq R_1\geq R_2\dots\)) and converges to \(0\).

In the language of uniform convergence (but see comments at the start of this post), it follows that the partial sum functions \(S_m\) converge to \(f\) uniformly on \(I.\)

Now that we have stated this theorem, the proof is fairly easy! I'll give the proof in Part II.