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Thread: Series Convergence

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

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

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

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

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

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

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

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

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