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Math Help - Series Convergence

  1. #1
    Senior Member I-Think's Avatar
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    Series Convergence

    Question
    Given a convergent series \sum{a_n}, prove this implies the convergence of \sum{\frac{\sqrt{a_n}}{n}} when a_n\geq{0}

    Proof
    I'm attempting a proof by contradiction
    So assume \sum{\frac{\sqrt{a_n}}{n}} divergent

    \sum{a_n} bounded, so \exists a least upper bound M

    {\frac{\sqrt{a_n}}{n}} divergent, so \exists n\in{\mathbb{N}} such that

    \sum_{i=1}^{n-1}{\frac{\sqrt{a_i}}{i}}<M and

    \sum_{i=1}^{n}{\frac{\sqrt{a_i}}{i}}>M

    So \frac{\sqrt{a_n}}{n}>\sum_{i=n+1}^{\infty} a_n

    And here I am stuck
    How do I continue, or should I try a new route?
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  2. #2
    Super Member girdav's Avatar
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    Re: Series Convergence

    For x,y\in\mathbb R we have xy\leq \frac{x^2+y^2}2.
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