What is the interval of convergence for the given sequence?

[xllGoliathllx]

I got a summer homework packet that covers all of last year's curriculum (Pre-Calc) and the first three chapters of next year's curriculum (Calc). I have no idea what I'm really doing with the calc stuff. I'm trying to teach myself sequences, limits, and derivatives. It was going pretty smoothly until this problem:

Find the interval of convergence for: 1 + 4x + 16x^2 + ...

A) x < 1/4
B) -1/4 < x < 1/4
C) 0< x < 1
D) x > -1/4

Could some please help me through this?

I'm sure this problem is ridiculously easy, but I can't seem to find anything to teach myself how to do it.

Sorry if I posted this in the wrong section.

 

[vela]

First, identify what type of series that is. You should know the condition under which that type of series converges. Apply that requirement to this problem and you'll have your answer.

 

[xllGoliathllx]

First, identify what type of series that is. You should know the condition under which that type of series converges. Apply that requirement to this problem and you'll have your answer.

One problem... I've never learned about series. From what I've read, I would say that is a power series. Is that right?

 

[vela]

That's weird. Usually you see series in algebra. Your series is a special case of a power series, called a geometric series.

http://en.wikipedia.org/wiki/Geometric_series

 

[xllGoliathllx]

That's weird. Usually you see series in algebra. Your series is a special case of a power series, called a geometric series.

http://en.wikipedia.org/wiki/Geometric_series

I might have seen it before and just forgot about it. Actually come to think of it, I learned about sequences, summation notation, etc last year. I've never seen some of these terms though. I'll take a look at that page and see if I can figure it out.

Thanks for the help,

Matt

 

[xllGoliathllx]

I think 0 < x < 1 makes the most sense. ?

 

[vela]

Nope. If you have a geometric series 1+r+r²+..., it converges if |r|<1. So what is r, the ratio between successive terms, for the series you've been given? Plug that into the inequality and solve for x.

 

[xllGoliathllx]

Nope. If you have a geometric series 1+r+r²+..., it converges if |r|<1. So what is r, the ratio between successive terms, for the series you've been given? Plug that into the inequality and solve for x.

Common ratio is 4.

Since the |4| is not less than 1, the series doesn't converge? I'm lost...

Sorry... I'm actually a pretty smart person when high school teachers actually teach before they give assignments.

It's just 1 question out of 140, so I don't care any more. You don't have to help me... I know it must feel like trying to teach a rock. lol

 

[vela]

Don't give up yet. It's pretty straightforward.

The series is 1+(4x)+(4x)²+... Compare that to the form of the generic geometric series 1+r+r²+... You can see that r=4x. So...

 

[xllGoliathllx]

Don't give up yet. It's pretty straightforward.

The series is 1+(4x)+(4x)²+... Compare that to the form of the generic geometric series 1+r+r²+... You can see that r=4x. So...

lol... I'm an idiot.

4x<1
x<1/4


Thanks again for the help :),

Matt

 

[vela]

Don't forget the absolute value! It's |r|<1, not r<1.

 

[HallsofIvy]

lol... I'm an idiot.

4x<1
x<1/4


Thanks again for the help :),

Matt

Don't forget the absolute value! It's |r|<1, not r<1.

And so the interval of convergence is
-\frac{1}{4}\le x&lt; \frac{1}{4}