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Interval of Convergence of a Series

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the interval of convergence for the following power series. Specify both absolute and conditional convergence where appropriate.


    2. Relevant equations
    1 + x + 2x^2 + 6x^3 + ... + n! x^n + ...


    3. The attempt at a solution
    Using the ratio test to determine convergence of the series, I have the following:

    (n+1)! x^(n+1)/[n! x^n] = (n!)(x^(n+1))/[(n!)(x^n)]
    = x(n+1) (After cancellation of factorials)

    = x * lim n->∞ (n+1)

    x < 1 converges
    x > 1 diverges
    Series does not converge?

    I'm not sure if my answer is right or not. Any help is much appreciated. Thanks!
     
  2. jcsd
  3. Aug 4, 2013 #2

    Zondrina

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    If you're asking if this converges or not :

    ##\sum_{n=0}^{∞} (n!)x^n## then yes you are correct, the series does not converge.

    Notice the terms are blowing up as n→∞.
     
  4. Aug 4, 2013 #3
    So what is ## \lim_{n \to \infty} | x(n + 1)| ##?
     
  5. Aug 4, 2013 #4
    It always goes to infinity regardless of the value of x, I think.
     
  6. Aug 4, 2013 #5
    How about x = 0?
     
  7. Aug 4, 2013 #6
    That would give you an indeterminate form of infinity times zero, I believe.
     
  8. Aug 4, 2013 #7
    Are you saying that the series ## 0 \cdot n ## has no particular limit as ## n \to \infty##?
     
  9. Aug 4, 2013 #8
    Well that is what I think of it as now, since it is undefined?
     
  10. Aug 4, 2013 #9

    micromass

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    But ##0\cdot n = 0## for all ##n##!
     
  11. Aug 4, 2013 #10
    Wait.. I thought infinity times zero is an indeterminate form, therefore that is undefined? Am I wrong in presuming that? So it doesn't matter, even if n is infinity- then n * 0= 0?! Wow, I was so misguided.
     
  12. Aug 4, 2013 #11

    micromass

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    Infinity times 0 is indeed undefined, but you're not doing infinity times zero here. You are taking the limit of ##0\times n##. But ##n## is always finite, so ##0\times n = 0## for all ##n##. Taking the limit of this yields ##0##.
     
  13. Aug 4, 2013 #12
    Infinity times zero indeterminacy arises when the "zero" is not really a zero, but something that converges to it, and the "infinity" is not really an infinity, but something that grows unbounded. THEN you cannot just say that it is zero or not, you need additional analysis.

    In this case, every member of the sequence is just zero, so nothing goes to zero nor infinity - you just have an infinite string of zeros.
     
  14. Aug 4, 2013 #13
    Thank you both for explaining that in detail. So in this case, then the sequence would converge at x=0 if I am not mistaken? Though I do not understand how that could be the case, if my calculations were correct.
     
  15. Aug 4, 2013 #14
    Not only can you conclude that the series converges at x = 0, but you can find the value it converges to. It is trivial.
     
  16. Aug 4, 2013 #15
    I hope I'm not overcomplicating things, but the value it converges to IS zero right? Because the problem I'm having here is how do you know if you can change the values of n AND x? If that is the case, then 0 * n= 0. But how do you know that you can do that?
     
  17. Aug 4, 2013 #16
    You do not change both x and n. When you test or prove the convergence, you assume that x is fixed.

    The series ## 0 \cdot n ## converges to zero. The original series converges to something else at x = 0, and you can easily find out its limit.
     
  18. Aug 4, 2013 #17
    I thought the original series does not converge at all though.
     
  19. Aug 4, 2013 #18
    Plug x = 0 into the original series. What do you get?
     
  20. Aug 4, 2013 #19
    I would get zero, I think, if n was held constant?
     
  21. Aug 4, 2013 #20
    You cannot hold n constant in a series. If you do, then you are talking about a member of a series, not of the entire series.

    So what is the sum of the original series when x = 0?
     
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