Power Series Question | Limit and Convergence | Solution Attempt

[Jbreezy]

Homework Statement

Question. Did I do this OK?

Homework Equations

The Attempt at a Solution

A_n = Ʃ e^(n^2) x^n from n = 1 to ∞

So I tried the root test. After you take the nth root you have x e^n so then I took the limit of this as n-->∞ and I got infinity. I then said OK |x| < 1 for this to converge so the interval of convergence is (-1,1) ?? This OK?

 

[LCKurtz]

Homework Statement

Question. Did I do this OK?

Homework Equations

The Attempt at a Solution

A_n = Ʃ e^(n^2) x^n from n = 1 to ∞

Why is there an $n$ on the left side and not the right side?

So I tried the root test. After you take the nth root you have x e^n so then I took the limit of this as n-->∞ and I got infinity. I then said OK |x| < 1 for this to converge so the interval of convergence is (-1,1) ?? This OK?

No, it isn't OK. It would help if you would write equations instead of prose. Show us your work instead of just declaring $|x|<1$.

 

[Jbreezy]

Why is there an $n$ on the left side and not the right side?

I don't see it sorry. Where?


No, it isn't OK. It would help if you would write equations instead of prose. Show us your work instead of just declaring $|x|<1$.

I don't know what prose is.
I just did
(e^(n^2) x^n)^(1/n) = e^n (x)
Took the limit as n --> infinity I got infinity.
This is when I said |x| < 1
So I got interval of convergence (-1,1)

 

[LCKurtz]

{A_n} = Ʃ e^(n^2) x^n from n = 1 to ∞

Why is there an $n$ on the left side and not the right side?

The $n$ is right there. And please don't put your question (where's the $n$) as part of my quote.

I don't know what prose is.

So look it up in a dictionary.

I just did
(e^(n^2) x^n)^(1/n) = e^n (x)
Took the limit as n --> infinity I got infinity.
This is when I said |x| < 1

I know that's what you said. It's wrong. Explain how you got |x| < 1.

 

[Jbreezy]

The $n$ is right there. And please don't put your question (where's the $n$) as part of my quote.



So look it up in a dictionary.



I know that's what you said. It's wrong. Explain how you got |x| < 1.

Didn't know it was in the quote. I got |x| <1 because by the root test it converges for those values. 0 < 1. It is what they did in my book when they got infinity for the limit. They took whatever was in abs like |x-3| or in this case |x| and did |x| <1 and found the interval of convergence from it.

 

[Mark44]

Didn't know it was in the quote. I got |x| <1 because by the root test it converges for those values.

Who is "it"?
Show us your work for the root test. That's what LCKurtz is asking for.

Regarding his comment about prose, what he's saying is more math, and less words.

0 < 1. It is what they did in my book when they got infinity for the limit. They took whatever was in abs like |x-3| or in this case |x| and did |x| <1 and found the interval of convergence from it.

 

[Jbreezy]

Who is "it"?
Show us your work for the root test. That's what LCKurtz is asking for.

Regarding his comment about prose, what he's saying is more math, and less words.

This is what I did

(A_n)^(1/n) = (e^(n^2) x^n)^(1/n) = e^n (x)

|(A_n)^(1/n)| = |x| lim n---> ∞ e^n = ∞ = L

OK the series diverges but if I want it to converge I need|x| < 1
So that's where it came from

 

[Mark44]

This is what I did

(A_n)^(1/n) = (e^(n^2) x^n)^(1/n) = e^n (x)

|(A_n)^(1/n)| = |x| lim n---> ∞ e^n = ∞ = L

OK the series diverges but if I want it to converge I need|x| < 1
So that's where it came from

So, for example, is
$$\sqrt[4]{e^{4^2}} = e^4?$$

 

[Jbreezy]

Yes.

 

[Mark44]

So you to find the values of x for which
$$|x|\lim_{n \to \infty}e^n < 1$$

|x| < 1 isn't going to cut it.

 

[Jbreezy]

So you to find the values of x for which
$$|x|\lim_{n \to \infty}e^n < 1$$

|x| < 1 isn't going to cut it.

I;m lost.

This is $$\lim_{n \to \infty}e^n = ∞ $$

This is $$|x| < 1 $$ for the series to converge? Yes. I swear my book has this. I don't understand why this is wrong. Is it not true?

 

[Ray Vickson]

I;m lost.

This is $$\lim_{n \to \infty}e^n = ∞ $$

This is $$|x| < 1 $$ for the series to converge? Yes. I swear my book has this. I don't understand why this is wrong. Is it not true?

No, if is as false as it is possible for anything to be! For example, suppose x = 1/10. Compute several of the partial sums
A_N = \sum_{n=1}^{N} e^{n^2} x^n for, say
N = 10, N = 20, N = 30. (Use a numerical computer package if you need to.) Does it look to you as if the values are converging to a finite limit?

Now try another x, such as x = 1/100. Do the same calculations and answer the same questions.

 

[vela]

Didn't know it was in the quote. I got |x| <1 because by the root test it converges for those values. 0 < 1. It is what they did in my book when they got infinity for the limit. They took whatever was in abs like |x-3| or in this case |x| and did |x| <1 and found the interval of convergence from it.

You've misunderstood what the book is saying, so the conclusion you're drawing based on your misconception isn't making sense to us. Try going over the example in the book again more carefully. If you're still confused, post the example from the book and explain to us what you think it's saying.