# Power Series and radius/interval of convergence

• Oct 8th 2009, 09:53 PM
Sam1111
Power Series and radius/interval of convergence
Hi

The problem is to find the radius and interval of convergence of the series...

((-1)^(k+1)*(x+1)^k)/k from k=1 to infinity.

I have attempted using the ratio test for absolute convergence and have simplified it down to

rho=lim(k->infinity) |(x+1)|
I know it will converge absolutely if rho is less than 1 so i got |(x+1)|<1

So is the radius of convergence is 1 and the interval of convergence -2<x<0?
• Oct 8th 2009, 10:25 PM
tonio
Quote:

Originally Posted by Sam1111
Hi

The problem is to find the radius and interval of convergence of the series...

((-1)^(k+1)*(x+1)^k)/k from k=1 to infinity.

I have attempted using the ratio test for absolute convergence and have simplified it down to

rho=lim(k->infinity) |(x+1)|
I know it will converge absolutely if rho is less than 1 so i got |(x+1)|<1

So is the radius of convergence is 1 and the interval of convergence -2<x<0?

Almost: the interval of convergence contains both 0 and -2.

Tonio
• Oct 8th 2009, 10:26 PM
Sam1111
Ok thanks! I was just making sure I was on the right track before checking the end points
• Oct 9th 2009, 06:26 AM
HallsofIvy
Quote:

Originally Posted by tonio
Almost: the interval of convergence contains both 0 and -2.

Tonio

That's not correct. If x= -2, the series is $\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k+1}(-1)^k}{k}= -\sum_{k=1}^\infty \frac{1}{k}$ which does not converge. The series converges for $\displaystyle -2< x\le 0$.