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Math Help - need help with Induction

  1. #1
    Junior Member
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    need help with Induction


    Sum of n=1 to N of 1/sqrt(n) > sqrt(N) for N>= 2.

    Thanks for any help.

    Last edited by alpha; January 19th 2010 at 08:29 AM.
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  2. #2
    Senior Member
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    Quote Originally Posted by alpha View Post
    Show using an inductive or non-inductive argument that \sum_{n=1}^N\frac{1}{\sqrt{n}} > \sqrt{N} for N\geq 2.
    Thanks for any help.
    Clearly N<N+1 for all N\geq 2. Then

    \sqrt{N}<\sqrt{N+1}

    \implies N<\sqrt{N}\sqrt{N+1}

    \implies N+1-\sqrt{N}\sqrt{N+1}<1

    \implies \sqrt{N+1}-\sqrt{N}<\frac{1}{\sqrt{N+1}}.

    So if \sqrt{N}<\sum_{n=1}^N\frac{1}{\sqrt{n}}, then

    \sqrt{N}+\sqrt{N+1}-\sqrt{N}<\sum_{n=1}^N\frac{1}{\sqrt{n}}+\frac{1}{\  sqrt{N+1}}

    \implies \sqrt{N+1}<\sum_{n=1}^{N+1}\frac{1}{\sqrt{n}}.

    Since furthermore \sqrt{2}<1+\frac{1}{\sqrt{2}}=\sum_{n=1}^2\frac{1}  {\sqrt{n}}, then \sum_{n=1}^N\frac{1}{\sqrt{n}} > \sqrt{N} for all N\geq 2.
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