Given a real power series sum_{n=0}^{+infty}a_n(x-x_0)^n+∞∑n=0an(x−x0)n, the radius of convergence is the quantity r = "sup" {tilde{r} \in \mathbb{R} : sum_{n=0}^{+infty}a_n tilde{r}^n " converges"}. Note that r >= 0, because for tilde{r}=0 the series sum_{n=0}^{+infty}a_n tilde{r}^n= sum_{n=0}^{+infty}a_n 0^n=1 converges (recall that 0^0=1).
This quantity it's a bound to the value taken by |x-x_0|. It's not hard to prove that the given power series will converge for every x such that |x-x_0| < r and it will not converge if |x-x_0|>r (the proof is based on the direct comparison test). The convergence of the case r=|x-x_0| depends on the specific power series.
This means that the interval (x_0-r,x_0+r) (the interval of convergence) is the interval of the values of x for which the series converges, and there are no other values of x for which this happens, except for the two endpoints x_{+}=x_0+r and x_{-}=x_0-r for which the convergence has to be tested case-by-case.
The term radius is thereby appropriate, because r describes the radius of an interval centered in x_0.
The definition of radius of convergence can also be extended to complex power series.
