The series...
... converges absolutely if , so that the circle of convergence has center in and radius ...
Kind regards
Use the ratio test: A series, , converges absolutely if the limit of the ratio [tex]|\frac{a_{n+1}}{a_n}|[/itex] is less than 1. For a power series, that ratio becomes . The "interval of convergence" is the interval in which x makes that less than 1.
In your example, and so that ratio is .
As n goes to infinity, [tex]\frac{n+1}{n}[tex] goes to 1 so that ratio goes to |x-2|. The series will converge if |x-2|< 1 or if -1< x- 2< 1 so that 1< x< 3. The radius of convergence is 1 and the interval of convergence is from 2-1= 1 to 2+1= 3.
That is on the real line. If x is allowed to be a complex number, then the "interval of convergence" is a disk in the complex plane with center 2 and radius 1. That is what Chisigma was referring to.
provided that 1 and 3 are correct (I didn't check any work, just looked at the post above), then yeah, you need to check the endpoints, so..
((-1)^n(x-2)^n)/n
plugging in 1..
[(-1)^n][(-1)^n]/n
or,
1/n
which diverges by the p-series test
and plugging in 3..
[(-1)^n][(1)^n]/n
or
[(-1)^n]/n
which converges by the alternating series test