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Math Help - Greatest integer of sum

  1. #1
    Senior Member pankaj's Avatar
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    Greatest integer of sum

    Let S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\  frac{1}{\sqrt{80}}

    Find [S] (where [S] denotes the greatest integer less than or equal to S)
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Let S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\  frac{1}{\sqrt{80}}

    Find [S] (where [S] denotes the greatest integer less than or equal to S)
    the answer is 16. the reason is the following inequality, which can be easily proved by either induction or routine calculus methods:

    \forall \ n \geq 2: \ \ 2 \sqrt{n+1} - 2 < \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n} - 1.
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    nevermind I bounded it by the wrong integral
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  4. #4
    MHF Contributor red_dog's Avatar
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    Use these inequalities:

    2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}, \ \forall n\geq 2

    for n=2, \ 3, \ \ldots,80
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  5. #5
    Senior Member pankaj's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    the answer is 16. the reason is the following inequality, which can be easily proved by either induction or routine calculus methods:

    \forall \ n \geq 2: \ \ 2 \sqrt{n+1} - 2 < \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n} - 1.

    How do we prove by calculus
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  6. #6
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    Krizalid's Avatar
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    you can actually prove that inequality by using calculus, thus, the rest follows.
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  7. #7
    Senior Member pankaj's Avatar
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    Actually I did it like this:
    \sqrt{n+1}>\sqrt{n}

    \sqrt{n+1}+\sqrt{n}>2\sqrt{n}

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

    \sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}

    Similarly,

    \sqrt{n-1}<\sqrt{n}

     <br />
\sqrt{n}+\sqrt{n-1}<2\sqrt{n}<br />

    \frac{1}{\sqrt{n}+\sqrt{n-1}}>\frac{1}{2\sqrt{n}}

    But how to do with Calculus

     <br />
\sqrt{n}-\sqrt{n-1}>\frac{1}{2\sqrt{n}}<br />
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  8. #8
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    for each k\in\mathbb N is \frac{1}{\sqrt{k+1}}<\int_{k}^{k+1}{\frac{dt}{\sqr  t{t}}}<\frac{1}{\sqrt{k}}.

    lower bound: 2\sqrt{n+1}-2=\int_{1}^{n+1}{\frac{dt}{\sqrt{t}}}=\sum\limits_  {k=1}^{n}{\int_{k}^{k+1}{\frac{dt}{\sqrt{t}}}}<\su  m\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}.

    upper bound: \sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}=1+\sum\l  imits_{k=1}^{n-1}{\frac{1}{\sqrt{k+1}}}<1+\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{\frac{dt}{\sqrt{t}}}}=1+2\sqrt{n  }-2=2\sqrt{n}-1.

    and we're done.
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