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Math Help - Deduce

  1. #1
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    Deduce

    Deduce that 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{  1}{\sqrt{n}}<2\sqrt{n}
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Deduce

    Since the question says deduce, I'm not sure if a solution that involves induction is acceptable.

    Anyway, using induction,

    We will need to prove: 2\sqrt{n}+\frac{1}{\sqrt{n+1}}<2\sqrt{n+1}

    \Leftrightarrow 2\sqrt{n}<\frac{2(n+1)-1}{\sqrt{n+1}}

    \Leftrightarrow 2\sqrt{n}\sqrt{n+1}<{2n+1}

    \Leftrightarrow 4n(n+1)<4n^2+4n+1

    \Leftrightarrow 0<1

    which is true.
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  3. #3
    Grand Panjandrum
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    Re: Deduce

    Quote Originally Posted by Punch View Post
    Deduce that 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{  1}{\sqrt{n}}<2\sqrt{n}
    You can to use:

    \sum_{k=1}^n \frac{1}{\sqrt{n}} < \int_1^{n+1}\frac{1}{\sqrt{x}}\; dx

    (but that is a calculus derivation)

    CB
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