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Math Help - Mathematical Induction

  1. #1
    Newbie
    Joined
    Apr 2007
    Posts
    15

    Mathematical Induction

    Hi, I started this problem, got stuck, and need a hint to continue.

    Prove that  1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{n^2} \leq <br />
2 - \frac{1}{n}

    Base case:
     \frac{1}{1^2} = 2 - 1
     1 =1 so the inequality holds for  n =1 .

    Induction Hypothesis:
    Assume that
     1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} \leq <br />
2 - \frac{1}{k} , where  k is an arbitrary integer  \geq 1.

    Prove that   1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{(k+1)^2} \leq <br />
2 - \frac{1}{k+1} <br />

    Here is what I got:
     1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{(k+1)^2} \leq <br />
2 - \frac{1}{k} + \frac{1}{(k+1)^2}

    Any suggestion on how to continue?
    Thanks in advanced.
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  2. #2
    MHF Contributor
    Joined
    Mar 2007
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    1,240

    Talking

    . . . . . \frac{1}{k}\, -\, \frac{1}{(k\, +\, 1)^2}\, =\, \frac{(k\, +\, 1)^2\, -\, k}{k(k\, +\, 1)^2}

    . . . . . =\, \frac{k^2\, +\, k\, +\, 1}{k(k\, +\, 1)^2}

    . . . . . >\, \frac{k^2\, +\, k}{k(k\, +\, 1)^2}\, =\, \frac{k(k\, +\, 1)}{k(k\, +\, 1)^2}

    . . . . . =\, \frac{1}{k\, +\, 1}

    Use the above and the fact that, if you subtract something smaller from 2, you'll end up with something larger, to prove the final inequality.

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