How would you find the series radius and interval of convergence? and for what value of x does the series converge absolutely, conditionally?
series n=1 to infinity ((-1)^n(x-2)^n)/n
The series...
$\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\cdot \frac{(x-2)^{n}}{n}$
... converges absolutely if $\displaystyle |x-2|<1$, so that the circle of convergence has center in $\displaystyle x=2$ and radius $\displaystyle r=1$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Use the ratio test: A series, $\displaystyle \sum a_n$, converges absolutely if the limit of the ratio [tex]|\frac{a_{n+1}}{a_n}|[/itex] is less than 1. For a power series, $\displaystyle \sum b_n(x-a)^n$ that ratio becomes $\displaystyle |\frac{b_{n+1}(x-a)^{n+1}}{b_n(x-a)^n}|= |\frac{b_{n+1}}{b_n}||x- a|$. The "interval of convergence" is the interval in which x makes that less than 1.
In your example, $\displaystyle b_n(x-a)^n= \frac{(-1)^n}{n}(x-2)^n$ and $\displaystyle b_{n+1}(x-a)^{n+1}= \frac{(-1)^{n+1}}{n+1}(x-2)^{n+1}$ so that ratio is $\displaystyle |\frac{(-1)^{n+1}(x-2)^{n+1}}{n+1}\frac{n}{(-1)^n(x-2)^n|= \frac{n+1}{n}|x-2|$.
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