TITLE: Why do we say "radius" of convergence?
QUESTION [10 upvotes]: In an intuitive sense, I have never understood why a power series centered on $c$ cannot converge for some interval like $(c-3,c+2]$.
Also, I have had a few professors casually mention that a series converges for a disk in the complex plane, centered on $c$ and with the radius of convergence as its radius.
Is this just a deep result that I havent seen yet because I havent taken enough real/complex analysis? Or is there an obvious reason for this.
Edit: perhaps a better way to ask my question: why is it that, if the series converges when $x$ is between $c$ and $c+a$, then it also converges for $x$ between $c$ and $c-a$

REPLY [7 votes]: The relevant theorem is the following.  
Theorem: Let $\sum a_nz^n$ be a complex power series.  Suppose that, for some $w\in\mathbb C$, the series $\sum a_nw^n$ converges.  Then for all $z\in\mathbb C$ such that $|z|<|w|$, the series $\sum a_nz^n$ converges absolutely.
Proof. Suppose that $\sum a_nw^n$ converges.  We shall in fact only need to use a fact that appears to be much weaker - that the individual terms $a_nw^n$ are bounded in modulus.  In other words, there exists some $M$ such that $|a_nw^n|\le M$ for all $n$.  
Now suppose $z\in\mathbb C$ and that $|z|<|w|$.  We shall show that the series $\mathbb a_nz^n$ converges absolutely; i.e., that the series $\sum |a_n||z|^n$ converges.  To show this, note that:
\begin{align}
|a_n||z|^n&=|a_nw^n|\left(\frac{|z|}{|w|}\right)^n \\
&\le M\left(\frac{|z|}{|w|}\right)^n
\end{align}
Now just use the fact that the (geometric) series $M\sum\left(\frac{|z|}{|w|}\right)^n$ converges (since $|z|<|w|$).  $\Box$
Can you see why this implies the following?
Corollary: Let $\sum a_nz^n$ be a complex power series.  Then either $\sum a_nz^n$ converges everywhere or there exists some $R\ge 0$ (the radius of convergence such that $\sum a_nz^n$ converges absolutely for $|z|<R$ and diverges for $|z|>R$.  (In general, we can say nothing about the case when $|z|=R$.)