Convergence – Intuitive Explanation of Proof of Abel’s Limit Theorem

Question

Assume the series
$$f(x)=\sum_{n=0}^{\infty}a_n x^n$$ converges for $-r<x<r$. Abel's theorem says that if the series also converges at $x=r$ then $\lim_{x\to r-} f(x)$ exists and we have
$$\lim_{x\to r-}f(x)=\sum_{n=0}^{\infty}a_n r^n$$. Moreover uniform convergence extends to $x=r$.

Every proof I've seen uses summation by parts and rewrites the series using either $c_n=\sum_{k=0}^n a_k$ or $A_n = \sum_{k=n}^{\infty}a_k$.

Is there a way to geometrically understand the proof of uniform convergence and continuity at $x=r$. What I mean is that if someone asked me to explain why summation by parts would work in the proof and why it would be related to the uniform convergence and continuity I would not have a good answer.

Answer 1

Since Abel's limit theorem can be proven using the Dirichlet convergence test (see these notes on Ken Davidson's webpage), perhaps you will be satisfied with a geometrically intuitive proof of the latter. I might edit this answer later to incorporate this step.

It is easy to spot that the Dirichlet test is a generalization of the "alternating series test". However, whereas the alternating series is proven "geometrically", the Dirichlet test is usually proven by an uninspiring application of the summation by parts formula. For motivation, let us briefly look at the alternating series test;

Alternating series test: Let $(a_n)_{n \geq 0}$ be a monotone sequence of positive reals with limit zero. Then, $\sum_{n=0}^\infty (-1)^n a_n$ converges. Indeed, the sequence $s_n = \sum_{i=0}^n (-1)^i a_i$ of partial sums satisfies $|s_m -s_n| \leq a_n$ for $m >n$ and is Cauchy, in particular.

This all follows from the fact that there is a nested sequence of closed intervals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \ldots$ with $\mathrm{length}(I_n) =a_n$ such that $s_n \in I_n$ for each $n$. Just take \begin{align*} I_0=[0,a_0] && I_1 = [s_1,s_0] && I_2 = [s_1,s_2] && I_3 = [s_3,s_2] && I_4 = [s_3,s_4] && \cdots \end{align*}

The plan is to generalize this approach to the setting of Dirichlet's test.

Dirichlet test: Let $(z_n)_{n \geq 0}$ be a sequence of complex numbers whose sequence of partial sums is bounded. So, there exists a closed disk $D$ of diameter $d$ such that $\zeta_n = \sum_{i=0}^n z_i \in D$ for all $n$. Let $(a_n)_{n \geq 0}$ be a monotone sequence of positive reals converging to zero. Then, $ \sum_{n=1}^\infty a_n z_n$ converges. Indeed, the sequence $s_n = \sum_{i=0}^n (-1)^i a_i$ of partial sums satisfies $|s_m -s_n| \leq da_n$ for $m >n$ and is Cauchy, in particular.

Remark: In the case $z_n=(-1)^n$, we can take $D=[0,1]$ so that $d=1$.

This will all follow once we construct a nested sequence of closed disks $D_0 \supseteq D_1 \supseteq D_2 \supseteq \ldots$ with $\mathrm{diam}(D_n) = da_n$ such that $s_n \in D_n$ for each $n$. The little piece of geometry we will use to accomplish this is the following:

Rescaling a disk with respect to a point other than the centre: Let $D$ be a closed disk in the plane. Consider a transformation $f(z) = a(z-z_0)+z_0$ where $a \in (0,1]$. This effects a scaling down by the factor $a$ with respect to the basepoint $z_0$. Suppose that $z_0$ belongs to the disk $D$ (but is not necessarily the centre). Then,

1. $f(D)$ is, again, a disk.
2. $f(D) \subseteq D$
3. the diameter of $f(D)$ is $a$ times the diameter of $D$.

A couple brief comments: (1) is quite intuitive--if you are standing somewhere inside a circle, it's "circleness" doesn't depend whether you measure distance in one unit system or another. (2) would be true replacing $D$ by any convex set (or even any set star-shaped about $z_0$).

Define $\zeta_n = \sum_{i=1}^n z_i$ so that, by assumption, $\zeta_n \in D$ for all $n$. Define transformations $f_n : \mathbb{C} \to \mathbb{C}$ by $$ f_n(z) = \frac{a_{n}}{a_{n-1}} (z-s_{n-1})+s_{n-1},$$ noting $0 < \frac{a_{n}}{a_{n-1}} \leq 1$. It is easy to check that:

• $s_0=a_0\zeta_0$
• $s_1=f_1(a_0\zeta_1)$
• $s_2=f_2 \circ f_1(a_0\zeta_2)$
• $s_n=f_n \circ \cdots \circ f_1(a_0\zeta_n)$, in general.

In particular \begin{align*} s_n \in D_n && \text{ where } && D_n = f_n \circ \cdots \circ f_1 (a_0D) \end{align*}