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Math Help - Bounded alternating series

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Bounded alternating series

    Does  \{a_n\} converging imply  \sum_{n=1}^\infty (-1)^na_n is bounded?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    No, take a_n=\frac{(-1)^n}{n}.

    It's true if a_n>0 though.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    No, take a_n=\frac{(-1)^n}{n}.

    It's true if a_n>0 though.
    Whoops, I meant to say  a_n>0 . Why is this true?
    Last edited by chiph588@; May 19th 2010 at 06:30 PM.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Whoops, I meant to say  a_n>0 . Why is this true?
    I'm sorry, it's false! Take a_n=1+\frac{(-1)^n}{n}...

    It's true if a_n>0 and converges monotonically.
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  5. #5
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    Quote Originally Posted by Bruno J. View Post
    I'm sorry, it's false! Take a_n=1+\frac{(-1)^n}{n}...

    It's true if a_n>0 and converges monotonically.

    Take a_n=2 - 1/n

    1/(n+1) < 1/n
    => 2 - 1/(n+1) > 2 - 1/n

    Here a_n>0 , is strictly monotone converging to 2.

    So, lim|a_n|=2 and not 0. For a convergent series we need that the tail end of the series go to 0.
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  6. #6
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    So it should be false , but can we find all the sequences that are the counterexamples for this statement ?
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  7. #7
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    I just gave you one such example.

    For any C > 0 ,

    take a_n=C-(1/n)

    (a_n) converges to C , however \sum_{n=1}^\infty(-1)^na_n is not convergent.
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by ques View Post
    I just gave you one such example.

    For any C > 0 ,

    take a_n=C-(1/n)

    (a_n) converges to C , however \sum_{n=1}^\infty(-1)^na_n is not convergent.
    I was asking whether the sum is bounded, not convergent.
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by ques View Post
    I just gave you one such example.

    For any C > 0 ,

    take a_n=C-(1/n)

    (a_n) converges to C , however \sum_{n=1}^\infty(-1)^na_n is not convergent.
    What chip said! The partial sums you give are bounded.

    To prove it, let b_n=a_n-a where a=\lim_{n\to \infty}a_n. By the alternating series test, \sum_{j=1}^\infty (-1)^jb_j converges, which immediately implies the partial sums \sum_{j=1}^m(-1)^ja_j are bounded.
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