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Math Help - Help with solving a proof by induction

  1. #1
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    Help with solving a proof by induction

    Problem: Prove that 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} > \frac{2n}{n + 1} where n \geq 2

    What I have so far:

    1) Prove for n = 2

    1 + \frac{1}{2} > \frac{2(2)}{2 + 1}

    1\frac{1}{2} > \frac{4}{3}

    1.5 > 1.333... which is true.

    2) Assume for k

    1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} > \frac{2k}{k + 1} assumed true

    3) Assume for k + 1

    1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} + \frac{1}{k + 1} > \frac{2(k + 1)}{(k + 1)+1} not sure what to do

    ---------------------------
    How can I continue proving this, because this is where I get stuck on actually proving the inequality is true via induction.

    Also, have I made any mistakes so far? If so, then what and how could I fix them.

    Thanks for any help, BG
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BG5965 View Post
    Problem: Prove that 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} > \frac{2n}{n + 1} where n \geq 2

    What I have so far:

    1) Prove for n = 2

    1 + \frac{1}{2} > \frac{2(2)}{2 + 1}

    1\frac{1}{2} > \frac{4}{3}

    1.5 > 1.333... which is true.

    2) Assume for k

    1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} > \frac{2k}{k + 1} assumed true

    3) Assume for k + 1

    1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} + \frac{1}{k + 1} > \frac{2(k + 1)}{(k + 1)+1} not sure what to do

    ---------------------------
    How can I continue proving this, because this is where I get stuck on actually proving the inequality is true via induction.

    Also, have I made any mistakes so far? If so, then what and how could I fix them.

    Thanks for any help, BG
    Having assumed for k, you have:

    1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} > \frac{2k}{k + 1}

    Then you need to prove for k+1 (not "assume"). What you can try is:

    1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} + \frac{1}{k + 1} > \frac{2k}{k + 1} + \frac 1 {k+1} = \frac{2k + 1}{k + 1}

    by using the induction hypothesis.

    Then "all you have to do" is prove that:
    \frac{2k + 1}{k + 1} \ge \frac{2(k + 1)}{k + 2}

    Not that I have a clue as to where to begin on that last bit at the moment, probably straightforward but I haven't had enough caffeine.
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  3. #3
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    \frac{2k+1}{k+1} = \frac{(k+1) + k}{k+1} = \frac{k+1}{k+1} + \frac{k}{k+1} = 1 + \frac{k}{k+1}

    \frac{2(k+1)}{k+2} = \frac{2k+2}{k+2} = \frac{(k+2)+k}{k+2} = \frac{k+2}{k+2} + \frac{k}{k+2} = 1 + \frac{k}{k+2}

    So we have \frac{2k+1}{k+1} = 1 + \frac{k}{k+1} \geq 1 + \frac{k}{k+2} = \frac{2(k+1)}{k+2}

    And we're done.

    Also, if you're having trouble understanding how proofs by induction work, you might want to read this: http://en.wikipedia.org/wiki/Mathematical_induction
    Last edited by Defunkt; September 12th 2009 at 07:00 AM.
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  4. #4
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    Thanks so much both of you, I finally get it now!
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