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Power series absolute convergence/ Taylor polynomial

  1. Aug 9, 2014 #1
    1.



    What if absolute convergence test gives the result of 'inconclusive' for a given power series?

    We need to use other tests to check convergence/divergence of the powerr series but the matter is even if comparison or integral test confirms the convergence of the power series, we don't know which of three condition (1. converges only for center a, 2. converges for all x, 3. converges within the radius R ) the power series comes under. Well, definitely 1st is valid since its a subset of the other two, but we cannot confirm 2nd ,3rd condition I think.

    We just know it converges/diverges. So.. definitely in this case 3nd condition is not true because it is valid only when there exists R.

    Then..is it automatically condition 1 or 2 always valid?
     
    Last edited: Aug 9, 2014
  2. jcsd
  3. Aug 14, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
  4. Aug 15, 2014 #3
    Ok. My wording could be confusing because I am not really good at English.


    If we do absolute convergence ratio test on a given series, and if the result is 1, the test result is always inconclusive.

    And for power series, we know the interval and the radius of convergence are relevant to the result of absolute convergence ratio test because abs[(a_n+1)/(a_n)] < 1 will give us a boundary. However, we cannot really say about the interval and radius of convergence if the ratio test gives number 1 because the test is inconclusive, thus there is no information about the boundary.

    In this case, as far as I learned, (there would be other methods but) we need to use integral test and comparison test. I assume x of the power series can be any number, so we can take that as a constant when we do integral test and comparison test(we take n as our variable, therefore the value of x in the power series is irrelevant I think).

    Although we can do this, these two tests will give the result that explains if the series converges or not only. We would not know the radius and interval of convergence because these information are primarily from the nature of absolute ratio test. Hence, the power series domain is not determined.

    My question is, "in this case is there any way we can find the domain of power series?"

    -> well presence of domain is always true if the power series converges.


    [-(extra question) Actually I do not know on what ground 1 is the boundary of convergence and divergence. It is the ratio between a prior term and the next term, so if it is smaller than 1, the term is decreasing when the series goes to infinity and I understand this long term behavior, yet just not sure how we can generalize the ratio smaller than 1 for convergence. Like, how we can generalize and determine fast enough rate of decreasing terms.]
     
    Last edited: Aug 15, 2014
  5. Aug 15, 2014 #4

    mathman

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    I'm not sure what you are asking, but I'll try. Let r be the radius of convergence of a power series in z (complex). Then |z| < r, series converges; |z| > r, series diverges. For |z| = r, the series must be examined in particular - there is no general rule.
     
  6. Aug 15, 2014 #5

    HallsofIvy

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    That can't happen. A power series has a specific "interval of convergence". The series converges absolutely inside the interval of convergence, diverges outside the interval of convergence. At the endpoints of the radius of convergence, the series may converge (but not absolutely) or diverge. The series has to be checked separately at the endpoints.

     
  7. Aug 16, 2014 #6

    mathman

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    A series may converge absolutely at the end points.
    example: ∑xn/n2 converges for |x| = 1, but diverges for |x| > 1.
     
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