# Math Help - Power Series

1. ## Power Series

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

2. The series...

$\sum_{n=1}^{\infty}(-1)^{n}\cdot \frac{(x-2)^{n}}{n}$

... converges absolutely if $|x-2|<1$, so that the circle of convergence has center in $x=2$ and radius $r=1$...

Kind regards

$\chi$ $\sigma$

3. Use the ratio test: A series, $\sum a_n$, converges absolutely if the limit of the ratio [tex]|\frac{a_{n+1}}{a_n}|[/itex] is less than 1. For a power series, $\sum b_n(x-a)^n$ that ratio becomes $|\frac{b_{n+1}(x-a)^{n+1}}{b_n(x-a)^n}|= |\frac{b_{n+1}}{b_n}||x- a|$. The "interval of convergence" is the interval in which x makes that less than 1.

In your example, $b_n(x-a)^n= \frac{(-1)^n}{n}(x-2)^n$ and $b_{n+1}(x-a)^{n+1}= \frac{(-1)^{n+1}}{n+1}(x-2)^{n+1}$ so that ratio is $|\frac{(-1)^{n+1}(x-2)^{n+1}}{n+1}\frac{n}{(-1)^n(x-2)^n|= \frac{n+1}{n}|x-2|$.

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.

4. Don't you have to test the end points for convergence too though since the interval of absolute convergence is finite? The end points for this problem are geometric right?

5. 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