Given a real power series #sum_{n=0}^{+infty}a_n(x-x_0)^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.
