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Thread: 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 $\displaystyle \{a_n\}$ be a sequence of positive real numbers such that $\displaystyle \sum a_n = \infty$. Show that

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

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

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

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

    Notation: $\displaystyle \Delta a_n = a_{n+1}-a_n$
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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 $\displaystyle x_n>0$ and $\displaystyle \sum {x_n}$ converges , then $\displaystyle \prod (1+x_n)$ converges also. Let $\displaystyle A_n = \sum_{k=0}^n a_k$. We have

    $\displaystyle \prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0$. Now since $\displaystyle A_n \to \infty$, we see that $\displaystyle \prod (1+\frac{a_k}{A_{k-1}})$ diverges, which implies that $\displaystyle \sum \frac{a_k}{A_{k-1}}$ diverges.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Very nice Paul!

    Here's my solution:

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

    $\displaystyle \prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0$. Now since $\displaystyle A_n \to \infty$, we see that $\displaystyle \prod (1+\frac{a_k}{A_{k-1}})$ diverges, which implies that $\displaystyle \sum \frac{a_k}{A_{k-1}}$ 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 $\displaystyle \{x_{n}\}_{n\in\mathbb{N}}$ be a sequence of nonnegative numbers, then
    $\displaystyle \sum_{n\in\mathbb{N}}x_{n}$ and $\displaystyle \prod_{n\in\mathbb{N}}(1+x_{n})$ 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; Mar 27th 2011 at 12:21 AM.
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