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Math Help - induction

  1. #1
    Junior Member
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    induction

    Show that for all positive integers n , we have.

    1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) < 2*sqrt(n)

    I have shown the base case:
    Now i need to assume the inductive case to show

    ...... 1/sqrt(n+1) < 2*sqrt(n+1)

    and there is when i get stuck.
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  2. #2
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    Apr 2009
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    Wink

    You now need to prove

    1/sqrt(n+1) < 2*sqrt(n+1)-2*sqrt(n)

    not the one written above

    If you still cannot solve this problem, please ask again.
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  3. #3
    Junior Member
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    Aug 2008
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    still dont get it

    I guess i dont know why I have to substract 2sqrt(n) from 2sqrt(n+1).
    nor do can i continue the induction after that
    thanks
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  4. #4
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    Wink

    The induction argument should go like this:

    Assuming

    1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) < 2*sqrt(n) (1) for some n

    We now prove

    1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) + 1/sqrt(n+1) < 2*sqrt(n+1) (2)

    Now if we can prove

    1/sqrt(n+1) < 2*sqrt(n+1)-2*sqrt(n) (3)

    When you add up (1) and (3) together, you will get (2).

    To prove (3), you can multiply both sides by sqrt(n+1), then complete the square.
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  5. #5
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    Florida
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    Use \frac{1}{\sqrt{n}}=\frac{2}{2\sqrt{n}}<\frac{2}{\s  qrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1}).

    Then when you sum you will have a telescoping series.
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