# Prove Abel's convergence theorem

In our analysis class today, our teacher wanted us to prove the following theorem, or according to him, known as Abel's theorem:

If $\sum\limits_{n=0}^{\infty} a_n (z-z_0)^n$ with $a_n \in \mathbb{C}$ converges at $z_1 \ne z_0$, then the power series converges for all $z$, such that $|z-z_0| < |z_1 - z_0|$.

I have a hard time understanding what this theorem is trying to say. Where did $z_1$ come from? And how will one be able to proof? I tried looking for this theorem online but I couldn't find anything.

Any help will be greatly appreciated. Thanks in advance!

• "... converges at some $z_1 \neq z_0$, then ..." – Amit Kumar Gupta Oct 30 '13 at 8:10
• This is not what I think of when I think of Abel's Theorem. It says that if $$A=\sum_{k=0}^\infty a_k$$ converges, and $$f(x)=\sum_{k=0}^\infty a_kx^k$$ then $$\lim_{x\to1}f(x)=A$$ if $x$ approaches $1$ within the wedge $|1-x|\le M(1-|x|)$. – robjohn Dec 10 '13 at 7:05

Here is the theorem in different words:

Consider a power series of the form $\sum_{n=0}^\infty a_n (z-z_0)^n$. Suppose that it converges at some point $z_1$ which is different from $z_0$. Then it converges at every point $z$ such that $|z-z_0|<|z_1-z_0|$.

The existence of $z_1$ is assumed. You should realize that a typical theorem consists of assumptions and conclusion(s). The assumptions here are:

• we have a power series of particular form
• it converges at $z_1$
• the inequality $|z-z_0|<|z_1-z_0|$ holds

The conclusion is:

• the series converges at $z$.

For the proof, you need to know three things:

1. A convergent series has bounded terms
2. Geometric series with ratio $<1$ converges
3. Comparison principle

Apply $1$ to $\sum_{n=0}^\infty a_n (z_1-z_0)^n$. Conclude that there is a number $M$ such that $|a_n||z_1-z_0|^n\le M$ for every $n$. Let $r=\frac{|z-z_0|}{|z_1-z_0|}$ and note that the series $\sum Mr^n$ converges by 2. Finally, use 3 and the inequality $|a_n(z-z_0)^n|\le Mr^n$.