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Math Help - Show that absolute sum converges

  1. #1
    Senior Member Dinkydoe's Avatar
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    Show that absolute sum converges

    Hoi,

    I was wondering if the following was true. I can't prove it...but it looks true

    suppose \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n| a_i-p| = \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |b_i-q| = 0

    for series in (a_i),(b_i) \subset \mathbb{R}

    Then \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n|a_i b_i -pq| = 0

    Sounds like something that should be trivially true..but i have hard time showing it..
    Last edited by Dinkydoe; October 14th 2012 at 10:11 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Show that absolute sum converges

    Quote Originally Posted by Dinkydoe View Post
    Sounds like something that should be trivially true..but i have hard time showing it..
    It is false. Choose for example a_i=b_i=1/i^2, then \sum_{i=1}^{\infty}a_i=\sum_{i=1}^{\infty}b_i=\pi^  2/6 but \sum_{i=1}^{\infty}a_ib_i=\sum_{i=1}^{\infty}1/i^4=\pi^4/90 and \pi^2/6\cdot \pi^2/6\neq \pi^2/90.


    P.S. I see that you have changed the hypothesis.
    Last edited by FernandoRevilla; October 14th 2012 at 10:20 AM.
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Re: Show that absolute sum converges

    yes sorry, I was editing it :/

    thanks for your reply though. It looks so convincing to me...and i need it for something. But i cant show it
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  4. #4
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    Re: Show that absolute sum converges

    There is a special case that is true with p,q = 0. Suppose that a_1 \leq a_2 \leq \cdots \leq a_n and b_1 \ge b_2 \ge \cdots \ge b_n. Then

    \frac{1}{n} \sum_{k=1}^n a_k b_k \leq \left(\frac{1}{n} \sum_{k=1}^n a_k\right) \left(\frac{1}{n} \sum_{k=1}^n b_k\right)

    directly by the Chebyshev inequality, and you are done.

    I have a feeling that your statement may be false in general but I'm having trouble finding counter-examples.
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  5. #5
    Senior Member Dinkydoe's Avatar
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    Re: Show that absolute sum converges

    I am thinking the following: it follows |a_i -p| \to 0 and |b_i -q|\to 0 implying that (a_i),(b_i) are Cauchy with limit p resp. q

    Then product |a_ib_i-pq|\to 0 as well.... (don't know the proof but product of 2 Cauchy series a_n\cdot b_n\to a\cdot b is something that should be true)

    If I'm correct then it should follow that 1/n\sum_{i=1}^n|a_ib_i-pq|\to 0

    which is the same as saying that a_k\to 0 implies 1/n \sum_{i=1}^n|a_i|\to 0

    Right? O.o
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  6. #6
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    Re: Show that absolute sum converges

    Actually, here's a proof.

    Let a_k = p+ c_k and b_k = q + d_k. Then you have that

    \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n |c_k| = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n |d_k| = 0

    Next,

    |a_kb_k-pq| = |c_k d_k+p d_k + q c_k| \leq |c_k d_k|+|p d_k| + |q c_k|

    Summing over k and dividing by n yields

    \frac{1}{n} \sum_{k=1}^n |a_kb_k-pq| \leq \frac{1}{n} \sum_{k=1}^n|c_k d_k|+\frac{1}{n} \sum_{k=1}^n|p d_k| + \frac{1}{n} \sum_{k=1}^n|q c_k|

    The latter two terms converge to 0 by what's given. We use the triangle inequality on the first term

    \left(\frac{1}{n} \sum_{k=1}^n|c_k d_k|\right)^2 \leq \left(\frac{1}{n} \sum_{k=1}^n| c_k|\right)\left(\frac{1}{n} \sum_{k=1}^n|d_k|\right) .

    This converges to 0 as well, so in the end, the sum on the LHS of the inequality also converges to 0.
    Thanks from Dinkydoe
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  7. #7
    Senior Member Dinkydoe's Avatar
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    Re: Show that absolute sum converges

    haha you genious. Thank you kindly
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  8. #8
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    Re: Show that absolute sum converges

    Haha, no problem. It's a cool question though
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