Question

|x-a| 14 years ago

Answer 1

If you have a power series
\[ \sum_{n = 0}^\infty a_n (x-a)^n\]
Via the ratio test, it converges iff
\[ \left| \frac{a_{n+1} (x-a)^{n+1}}{a_n (x-a)^n}\right| < 1 \]
This puts the condition on x-a that
\[|x-a| < \left| \frac{a_n}{a_{n+1}} \right| = R\]

Answer 2

|x-a|<R means -R<x-a<R or -R+a<x<R+a
In a Taylor series context, a should stand for point around which you discuss the series while R should stand for the radius of the development
The theory says that the series (as a function series) uniformly converges on any compact interval [c,d] inside (a-R,a+R) and diverges outside (a-R, a+R)
You have to study for each case the behavior on the boundary, i.e. x=a-R and x=a+R
The interval (a-R, a+R) with the boundaries modified by the above study gives you the convergence domain