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Math Help - Divergence of a series

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Divergence of a series

    Here's a challenge problem I just made up:

    Let \{a_n\} be a sequence of positive real numbers such that \sum a_n = \infty. Show that

    \frac{a_1}{a_0}+\frac{a_2}{a_0+a_1}+\frac{a_3}{a_0  +a_1+a_2}+ \dots = \infty.
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  2. #2
    Super Member PaulRS's Avatar
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    Let S_n = \sum_{k=0}^{n}a_k.

    We want to show that  \sum_{k=0}^n\frac{\Delta S_k}{S_k}} diverges as n\to \infty.
    ( I chose this notation to suggest the similarity with \int_{ [ a,\infty)} \tfrac{f'(x)}{f(x)}dx = +\infty , if \displaystyle\lim_{x\to\infty} f(x) = +\infty )

    We assume that \frac{\Delta S_k}{S_k}} \to 0 otherwise the series would diverge trivially.

    Let L_k = \log S_k, then \Delta L_k = \log S_{k+1} - \log S_k = \log \left( 1 +\frac{\Delta S_k}{S_k} \right) =\frac{\Delta S_k}{S_k} + O\left( \left(\frac{\Delta S_k}{S_k}\right)^2 \right)

    We observe that \sum_{k=0}^n \Delta L_k = \log S_{n+1} - \log S_0 diverges since \log S_n \to \infty and now by the limit comparison test - both differences are always positive-, we get that  \sum_{k=0}^n\frac{\Delta S_k}{S_k}} diverges too.

    Notation: \Delta a_n = a_{n+1}-a_n
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Very nice Paul!

    Here's my solution:

    It's well known that if x_n>0 and \sum {x_n} converges , then \prod (1+x_n) converges also. Let A_n = \sum_{k=0}^n a_k. We have

    \prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0. Now since A_n \to \infty, we see that \prod (1+\frac{a_k}{A_{k-1}}) diverges, which implies that \sum \frac{a_k}{A_{k-1}} diverges.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Lightbulb

    Quote Originally Posted by Bruno J. View Post
    Very nice Paul!

    Here's my solution:

    It's well known that if x_n>0 and \sum {x_n} converges , then \prod (1+x_n) converges also. Let A_n = \sum_{k=0}^n a_k. We have

    \prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0. Now since A_n \to \infty, we see that \prod (1+\frac{a_k}{A_{k-1}}) diverges, which implies that \sum \frac{a_k}{A_{k-1}} diverges.
    Your solution is same as what came to my mind when I saw the question.
    For those who might want to see a reference for the relation between the sum and the product used by Bruno J, please see [Theorem 7.4.6, 1].
    I quote the result for convenience.
    Lemma. Let \{x_{n}\}_{n\in\mathbb{N}} be a sequence of nonnegative numbers, then
    \sum_{n\in\mathbb{N}}x_{n} and \prod_{n\in\mathbb{N}}(1+x_{n}) converge or diverge together.

    References
    [1] L.S. Hahn and B. Epstein, Classical Complex Analysis, Jones and Bartlett Publishers, Inc., London, 1996.
    Last edited by bkarpuz; March 27th 2011 at 12:21 AM.
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