Why do we say “radius” of convergence?

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 $c-a$

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While it is true that in complex analysis, power series converges on discs (hence the name ‘radius of convergence’), this is not necessary to see why real power series converge on a symmetric interval about their centre.

A power series with real coefficients centred at the point $c$ can be written as
\sum_{n=0}^{\infty} a_n (x-c)^n,
and it will converge whenever we plug in a value of $x$ so that the resulting series of real numbers converges. If $b = |x-c|$, then we’re really examining the convergence of the series of numbers
\sum_{n=0}^{\infty} a_n b^n.
(This statement is precisely the ratio test!)
This will converge (say) when $0 \leq b < R$, so we are allowed to substitute in values of $x$ satisfying $0 \leq |x-c| < R$. Therefore, the values of $x$ for which the series $\sum_{n=0}^{\infty} a_n (x-c)^n$ are those in the interval $(c-R, c+ R)$. (One must, of course, check what happens when $x = c- R$ or $x = c+R$, as we could have convergence at one or both of the endpoints, but in any case we still get a symmetric interval about $c$.)

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:
|a_n||z|^n&=|a_nw^n|\left(\frac{|z|}{|w|}\right)^n \\
&\le M\left(\frac{|z|}{|w|}\right)^n
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 $|z|=R$.)

In complex analysis, the radius of convergence is an actual radius.
In real analysis we only have one axis, so the radii look like intervals.
Draw this on paper.

(c-3, c+ 2] has midpoint (c-3+ c+2)/2= c- 1/2. It can be written as ((c-1/2)-5/2, (c- 1/2)+ 5/2] so has “radius” 5/2.