Calculating the Radius of Convergence for a Series Using the Ratio Test

[BoundByAxioms]

Homework Statement

Find the radius of convergence for $$\Sigma$$ $$\frac{nx^{2n}}{2^{n}}$$

Homework Equations

Ratio test

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?

 

[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.

 

[BoundByAxioms]

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.

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

 

[Lamoid]

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

 

[BoundByAxioms]

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

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

 

[Lamoid]

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

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

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

 

[BoundByAxioms]

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

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

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

 

[BoundByAxioms]

This is why I don't stay up to 3 AM. I don't know why I said to take the root. You would only do that if you wanted the interval of convergence. You don't have to take the root. You just want to leave it as absolute x^2 < 2. Thus you say the radius of C is 2.

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

 

[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.

 

[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 $|x|< \sqrt{2}$.

 

[BoundByAxioms]

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 $|x|< \sqrt{2}$.

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.

 

[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.

 

[BoundByAxioms]

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.

Ok great! Thank you for the clarification.

 

[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 $\sqrt{2}$

 

[BoundByAxioms]

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 $\sqrt{2}$

That makes sense. Thanks.