Power Series Question | Limit and Convergence | Solution Attempt

In summary, the conversation was about determining the interval of convergence for a series using the root test. The person asking the question had used the root test and came to the conclusion that the interval of convergence was (-1,1), but others in the conversation pointed out that this was incorrect and asked for more explanation on how the person reached that conclusion. The person also mentioned their book using the same method for determining the interval of convergence, but others pointed out that they may have misunderstood the example in the book.
  • #1
Jbreezy
582
0

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?
 
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  • #2
Jbreezy said:

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##.
 
  • #3
LCKurtz said:
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)
 
  • #4
{A_n} = Ʃ e^(n^2) x^n from n = 1 to ∞

LCKurtz said:
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.

Jbreezy said:
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.
 
  • #5
LCKurtz said:
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.
 
  • #6
Jbreezy said:
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.
Jbreezy said:
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.
 
  • #7
Mark44 said:
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
 
  • #8
Jbreezy said:
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
$$\sqrt[4]{e^{4^2}} = e^4?$$
 
  • #9
Yes.
 
  • #10
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.
 
  • #11
Mark44 said:
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?
 
  • #12
Jbreezy said:
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
[tex] A_N = \sum_{n=1}^{N} e^{n^2} x^n[/tex] 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.
 
  • #13
Jbreezy said:
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.
 

FAQ: Power Series Question | Limit and Convergence | Solution Attempt

What is a power series?

A power series is an infinite series of the form ∑n=0∞ an(x-c)n, where an is a coefficient and c is a constant. It is a type of mathematical series that is useful for representing functions as an infinite sum of polynomials.

What is the convergence of a power series?

The convergence of a power series refers to the values of x for which the series converges, or approaches a finite value. This can be determined by using the ratio test or the root test, which evaluate the limit of the ratio or root of the terms in the series as n approaches infinity.

How can power series be used to approximate functions?

Power series can be used to approximate functions by truncating the series at a certain point and using the remaining terms to approximate the function. This is particularly useful for functions that are difficult to evaluate or integrate analytically.

What is the radius of convergence of a power series?

The radius of convergence of a power series is the distance from the center point c at which the series will converge. It is determined by the convergence tests and can be a finite or infinite value.

How are power series used in real-world applications?

Power series have many practical applications, such as in physics, engineering, and economics. They can be used to model various phenomena, such as motion, electromagnetic fields, and financial data. They are also used in numerical methods for solving differential equations and other mathematical problems.

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