# Homework Help: Radius of convergence

1. Aug 10, 2008

### BoundByAxioms

1. The problem statement, all variables and given/known data
Find the radius of convergence for $$\Sigma$$ $$\frac{nx^{2n}}{2^{n}}$$

2. Relevant equations
Ratio test

3. The attempt at a solution

I apply the ratio test to get $$\frac{(n+1)(x^{2})}{2n}$$. I let n approach infinity, to get $$\frac{1}{2}$$. So, this series converges when |x2|<1. So, I have -1<x2<1. Now I'm lost, where can I go from here?

2. Aug 10, 2008

### Lamoid

I think that should be x^2 < 2 actually. Since it will converge if x^2 / 2 is less than 1.Also, I don't believe you need to make it absolute value of x^2 since x^2 will always be positive.

Last edited: Aug 10, 2008
3. Aug 10, 2008

### BoundByAxioms

So it converges on this interval (-$$\infty$$,2)?

4. Aug 10, 2008

### Lamoid

Remember you have x^2 < 2. Just solve it for x. And about that absolute value thing, disregard what I said.

5. Aug 10, 2008

### BoundByAxioms

So, I get x<$$\pm$$$$\sqrt{2}$$ (I know that doesn't make sense).

6. Aug 10, 2008

### Lamoid

Ok, well my instruction in my first post to remove the absolute value signs was dumb. If you take the root of both sides you get root of x^2 < root of 2. Remember that the root of x^2 is the absolute value of x anyway. So you see why what I told you was such a waste of time?

To clarify, you have:

$$\left|x^{2}\right|<2$$ then

$$\left|x\right|<\sqrt{2}$$

Last edited: Aug 10, 2008
7. Aug 10, 2008

### BoundByAxioms

Ok I'm feeling pretty dense right now, so I'm going to say the radius is -$$\sqrt{2}$$<x<$$\sqrt{2}$$?

8. Aug 10, 2008

### BoundByAxioms

Haha, ok thank you! I see your an east coaster, lol. For me it's only 12.

9. Aug 10, 2008

### Lamoid

Just to clarify, I have no idea what I am talking about and should stop clarifying. I was right originally! It SHOULD be absolute x is less than root 2. You DO need need the radius to be in terms of x to the power of one. Live and learn. My apologies, I am watching a debate with creationists I could send you if you require me to make ammends.

10. Aug 10, 2008

### HallsofIvy

The instruction to take the root originally was correct. That is because the "2" in the exponent of x.

By the ratio test we must have
$$\frac{(n+1)|x|^{2(n+1)}}{2^{n+1}}\frac{2^n}{n|x|^n}= \frac{n+1}{n}\cdot\frac{1}{2}|x|^n< 1$$

The limit of that is not "1/2", it is (1/2)|x|2 which must be less than 1: (1/2)|x|2< 1 so |x|2< 2 and $|x|< \sqrt{2}$.

11. Aug 10, 2008

### BoundByAxioms

Ok thank you. So because it is a power series, it will converge when $$\frac{1}{2}$$|x2|<1, and diverge when $$\frac{1}{2}$$|x2|>1.

Last edited: Aug 10, 2008
12. Aug 10, 2008

### HallsofIvy

More correctly, it will converge absolutely for for (1/2)x2|< 1, diverge for (1/2)x2|> 1 and may converge conditionally or diverge when (1/2)x2|= 1.

That is, of course, because the ratio test says that a sum, $\sum_{n=0}^\infty a_n$ of positive numbers will converge if $lim a_{n+1}/a_n< 1$ converge for $lim a_{n+1}/a_n >1$ and may converge or diverge if that limit is 1.

13. Aug 10, 2008

### BoundByAxioms

Ok great! Thank you for the clarification.

14. Aug 11, 2008

### HallsofIvy

And, by the way, "the radius is $$-\sqrt{2}< x< \sqrt{2}$$" is wrong. That is the interval of convergence. The radius of convergence is $\sqrt{2}$

15. Aug 11, 2008

### BoundByAxioms

That makes sense. Thanks.