Radius of convergence (power series) problem

[aero_zeppelin]

Homework Statement

Ʃ (from n=1 to ∞) (4x-1)^2n / (n^2)
Find the radius and interval of convergence

The Attempt at a Solution

I managed to do the ratio test and get to this point:

| (4x-1)^2 |< 1

But now what? How do you get the radius and interval? Any help will be appreciated!

Thanks

 

[Dick]

Homework Statement

Ʃ (from n=1 to ∞) (4x-1)^2n / (n^2)
Find the radius and interval of convergence

The Attempt at a Solution

I managed to do the ratio test and get to this point:

| (4x-1)^2 |< 1

But now what? How do you get the radius and interval? Any help will be appreciated!

Thanks

You 'solve' the inequality. Figure out what values of x work. Try getting started. Here's a hint. |a|<1 means -1<a<1.

 

[aero_zeppelin]

Ok, so:

a) (4x-1)^2 will never be negative so you only focus on (4x -1)^2 < 1

b) I developed the squared binomial, and solved for x.

I got x < 1/2 , so...

Radius = 1/2 ?

Interval of Conv. = (- INF, 1/2] ?

 

[Dick]

Ok, so:

a) (4x-1)^2 will never be negative so you only focus on (4x -1)^2 < 1

b) I developed the squared binomial, and solved for x.

I got x < 1/2 , so...

Radius = 1/2 ?

Interval of Conv. = (- INF, 1/2] ?

x=(-1) doesn't work. Most quadratics have two roots.

 

[aero_zeppelin]

Ok. So is my solution correct or did I miss anything?

Thanks for the help btw!

 

[Dick]

Ok. So is my solution correct or did I miss anything?

Thanks for the help btw!

Well, no it's not correct (4x-1)^2=1 has two solutions. You missed one. 16x^2-8x=0 has two roots. If you divided by x you are going to miss one. What is it?

 

[aero_zeppelin]

Hmmm..

I factored out 16x^2 - 8x < 0 into 8x(2x -1) < 0

So, one solution is x < 0 , second solution x < 1/2 ?

 

[Dick]

Hmmm..

I factored out 16x^2 - 8x < 0 into 8x(2x -1) < 0

So, one solution is x < 0 , second solution x < 1/2 ?

No. You can't 'factor' inequalities like that. ab<0 doesn't mean a<0 or b<0. If (4x-1)^2=1 only at 0 and 1/2, then the intervals that can be in your solution set to the inequality are (-infinity,0), (0,1/2) or (1/2,infinity). Which of those intervals work? Pick a point in each one and test it, if you have to.

 

[aero_zeppelin]

What I meant is that the two solutions for the inequality are x < 0 and x < 1/2

Assuming those are right, then I try both points in the series and both converge, so the interval would be
(-INF, 1/2] or (-infinity,0][0,1/2] as you said.

That's all I got, I have no clue about the radius

 

[Dick]

What I meant is that the two solutions for the inequality are x < 0 and x < 1/2

Assuming those are right, then I try both points in the series and both converge, so the interval would be
(-INF, 1/2] or (-infinity,0][0,1/2] as you said.

That's all I got, I have no clue about the radius

The interval is NOT (-INF,1/2]. x=(-1) is in that interval. (4*(-1)-1)^2=(-5)^2=25>1! So (4x-1)^2<1 is NOT true if x=(-1). I'm not sure how you are going wrong exactly, but the correct interval of convergence is [0,1/2], since as you say, it converges at the endpoints. What's the radius of the interval [0,1/2]? The center of the interval is x=1/4.