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Math Help - Divergent fractional series

  1. #1
    Senior Member I-Think's Avatar
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    Divergent fractional series

    Given that \sum{a_n} diverges, and a_n\geq{0}, prove \sum{\frac{a_n}{a_n+1} diverges

    So for any given M\in{\mathbb{R}}, \exists N\in{\mathbb{N}} such that when n\geq{N}, \sum^n_{k=1}{a_k}\geq{M}

    Where do we go from here?
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  2. #2
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    Re: Divergent fractional series

    Quote Originally Posted by I-Think View Post
    Given that \sum{a_n} diverges, and a_n\geq{0}, prove \sum{\frac{a_n}{a_n+1} diverges
    So for any given M\in{\mathbb{R}}, \exists N\in{\mathbb{N}} such that when n\geq{N}, \sum^n_{k=1}{a_k}\geq{M} Where do we go from here?
    You have two cases: (a_n)\not\to 0 and (a_n)\to 0.

    Consider the implications of each case.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Divergent fractional series

    If \lim_{n \rightarrow \infty} a_{n} \ne 0 , then also \lim_{n \rightarrow \infty} \frac{a_{n}}{1+a_{n}} \ne 0 and the series \sum_{n} \frac{a_{n}}{1+a_{n}} diverges. If \lim_{n \rightarrow \infty} a_{n} = 0 , then for n 'large enough' is \frac{a_{n}}{1+a_{n}}> \frac{a_{n}}{2} and, because the series \sum_{n} \frac{a_{n}}{2} diverges, also the series \sum_{n} \frac{a_{n}}{1+a_{n}} diverges...

    Kind regards

    \chi \sigma
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  4. #4
    Senior Member I-Think's Avatar
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    Re: Divergent fractional series

    Okay, contemplated on hint, believe I got a solution

    For a_n\rightarrow{x\neq{0}}, \frac{a_n}{1+a_n}\rightarrow{y\neq{0}}, so result follows

    For a_n\rightarrow{0}, assume \sum\frac{a_n}{1+a_n} convergent

    For any given n, let Z=max(1+a_i), 1\leq{i}\leq{n}

    Choose M so \frac{M}{Z}>LUB(\sum\frac{a_n}{1+a_n}), choose N so that for n\geq{N}, \sum{a_n}>M

    Now \sum{a_n}<Z\sum\frac{a_n}{1+a_n}<M

    But \sum{a_n}>M, so a contradiction has occurred. The result follows

    Is this proof satisfactory?
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  5. #5
    Senior Member I-Think's Avatar
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    Re: Divergent fractional series

    Just saw chisigma's proof, it's slicker than mine, but I believe this is still correct, no?
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