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Absolute Convergent Series proof

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Let Sumation "a sub n" be an absolutely convergent series, and "b sub n" a bounded sequence. Prove that sumation "a sub n"*"b sub n" is convergent. (sorry fist time on this site and can't use the notation.)


    2. Relevant equations
    Theorem A: that states sumation "a sub n" = A and sumation "b sub n" = B. Then
    1. Sumation ("a sub n" + "b sub n") = A + B
    2. For each k in Real, sumation (k*"a sub n") = k sumation "a sub n" if when k does not = 0

    Theorem B: if sumation "a sub n" converges absolutely then sumation "a sub n" converges

    3. The attempt at a solution

    Since "b sum n" is bounded there exist a real number M that is the bound. I then use M as a constant and bring it out of the sumation so that we have "M * sumation "a sub n""
    We are now left with the the original absolutly convergent series times M. I know M must be greater than or = to 0 by definition of bounded. So the series will converge at ever point except 0. 0 can not be used .

    Please give some direction to my maddness. Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2007 #2

    Dick

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    A series is absolutely convergent if the absolute values of the series also converge. Can you show this for the product series?
     
  4. Oct 24, 2007 #3
    M times the absolutely converging sum is CONVERGENT as you rightly asserted. Factor the M out of the sum, now M is bigger than what?

    use the comparison test.
     
  5. Oct 25, 2007 #4
    I found a case were the summation of "a sub n" is absolute convergent times a "b sub n" that is bounded does not converge

    (Sumation 1/n^2) (-1)^n

    This product is bounded but does not converge.

    Do I just show this example as proof that original question is false?
     
  6. Oct 25, 2007 #5

    Dick

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    I admire your spirit. Trying to show something that is true is false hones your skills. But sum((-1)^n/n^2) is convergent.
     
  7. Oct 25, 2007 #6
    Is the following true?

    (any sequence)(any series) make the sequence part of the series?
     
  8. Oct 25, 2007 #7

    Dick

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    That doesn't make much sense. A sequence is a series of numbers with no summation. A series is a sequence of numbers summed. If you mean by sequence a_n time a series b_n is sum(a_n*b_n), then that's sort of ok, but you have to clearer about what you mean.
     
  9. Oct 25, 2007 #8
    Let Sum a_n converge absolulty
    Let b_n be bounded
    w.w.t.s. (we wish to show)
    A) (Sum a_n) *(b_n) converges

    My case was
    a_n =1/n^2
    b_n =(-1)^n
    B) As (sum (a_n*b_n)) it converges
    As b_n(sum a_n) it diverges due to the alternating -1 being outside the sum

    When given A (w/ or w/o convergent condition) do we always go to B?

    Thank you for your help as you can tell I really need it.
     
  10. Oct 26, 2007 #9
    like dick said, you have to be clear by what you mean when you are multiplying a series with a sequence. The way to make SENSE out of multiplying a series with a sequence is as follows, the SUM of a sub n is a sequence summed up, that is
    a_1 + a_2 +a_3....
    Now when we "multiply" the sequence b sub n we get
    (a_1 * b_1) + (a_2)(b_2)+ (a_3)(b_3) +....

    you get it?
    Now if you are multplying alternating (-1)^n with the sum of 1/n^2 you have a new series which is the same series 1/n^2 except now it's odd terms are negative and it's even terms are positive, but if we take the absolute VALUE of this new series we get the original one, which converges!

    write the first 5 partial sum of your supposed counterexample to see this.
     
  11. Oct 27, 2007 #10
    I understand the multiplying a series with a sequence now.

    Back to answering the problem

    We know
    b_n is bounded IFF there exist M s.t. b_n ≤ M
    If ∑a_n is absolutely convergent, then ∑a_n is convergent

    ∑a_n*b_n ≤ ∑a_n*M
    ∑a_n*b_n ≤ M∑a_n
    Since ∑a_n convergers then M∑a_n converges also
    By comperison ∑a_n*b_n ≤ M∑a_n
    Thus ∑a_n*b_n converges
     
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