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Proof for inequality induction

  1. Oct 17, 2007 #1
    proof for inequality induction...please help!!

    1. The problem statement, all variables and given/known data

    Prove for all positive integers n that
    [tex]\sum^{n}_{l=1} l^{-1/2} > 2(\sqrt{n+1} -1)[/tex]

    2. The attempt at a solution
    [tex]\sum^{n+1}_{l=1} l^{-1/2}
    = \sum^{n}_{l=1} l^{-1/2} + (n+1)^{-1/2} > 2(\sqrt{n+1} -1) + (n+1)^{-1/2} [/tex]

    please help meee! i'm getting stuck, on how to express [tex] 2(\sqrt{n+1} -1) + (n+1)^{-1/2} [/tex] in terms of [tex] 2(\sqrt{n+2} -1) [/tex]
    Last edited: Oct 17, 2007
  2. jcsd
  3. Oct 17, 2007 #2


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    Are you sure you are supposed to prove this by induction? I'm having a hard time doing it that way. On the other hand there is an easy way to do it by looking at an upper Riemann sum and comparing it with the integral of x^(-1/2).
  4. Oct 17, 2007 #3
    yes im pretty sureee..because that's what im learning right nowww :frown:
  5. Oct 17, 2007 #4


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    Ok, so what you want to do is now show that 2*(sqrt(n+1)-1)+1/sqrt(n+1)>2*(sqrt(n+2)-1). Right? Multiply both sides by sqrt(n+1). The simple sqrt(n+1) radicals cancel on both sides. The sqrt(n+1)sqrt(n+2) remains on the right. Square both sides. It just barely works.
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