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Homework Help: Power series question

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Question. Did I do this OK?

    2. Relevant equations

    3. 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?
  2. jcsd
  3. Dec 3, 2013 #2


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

    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##.
  4. Dec 3, 2013 #3
    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)
  5. Dec 3, 2013 #4


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    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.
  6. Dec 3, 2013 #5

    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.
  7. Dec 3, 2013 #6


    Staff: Mentor

    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.
  8. Dec 3, 2013 #7
    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
  9. Dec 3, 2013 #8


    Staff: Mentor

    So, for example, is
    $$\sqrt[4]{e^{4^2}} = e^4?$$
  10. Dec 3, 2013 #9
  11. Dec 3, 2013 #10


    Staff: Mentor

    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.
  12. Dec 3, 2013 #11

    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?
  13. Dec 3, 2013 #12

    Ray Vickson

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    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.
  14. Dec 4, 2013 #13


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    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.
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