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Thread: Inequality

  1. #1
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    Inequality

    Can somebody demonstrate:
    $\displaystyle \
    \frac{n}{{\sqrt[n]{{n!}}}} < \left( {1 + \frac{1}{n}} \right)^n
    \
    $
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  2. #2
    Super Member PaulRS's Avatar
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    First we write the inequality in the equivalent form: $\displaystyle
    \log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1}
    {n}} \right) < \tfrac{1}
    {n} \cdot \log \left( {n!} \right)
    $

    Note that $\displaystyle
    \log \left( {1 + \tfrac{1}
    {n}} \right) \geq \tfrac{1}
    {n} - \tfrac{1}
    {2} \cdot \left( {\tfrac{1}
    {n}} \right)^2
    $ - think about the Taylor series, or you can also prove that $\displaystyle \log(1+x)\geq{x-\tfrac{x^2}{2}}$ (x>0) by differentiating -

    Hence we have: $\displaystyle LHS=
    \log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1}
    {n}} \right) \leq \log \left( n \right) - 1 + \tfrac{1}
    {{2n}}
    $

    Next note that: $\displaystyle
    \log \left( {n!} \right) = \sum\limits_{k = 1}^n {\log \left( k \right)} = \int_0^n {\log \left( x \right)dx} + \int_0^n {\tfrac{{\left\{ x \right\}}}
    {x}dx}
    $ - see Lemma 1, here-

    So, since $\displaystyle
    \int_0^n {\tfrac{{\left\{ x \right\}}}
    {x}dx} \geqslant \int_0^1 {\tfrac{{\left\{ x \right\}}}
    {x}dx} = \int_0^1 {dx} = 1
    $ we find: $\displaystyle
    \log \left( {n!} \right) \geq n \cdot \log \left( n \right)-n + 1
    $

    Hence: $\displaystyle
    \log \left( n \right) - 1 + \tfrac{1}
    {n} \leq \tfrac{1}
    {n} \cdot \log \left( {n!} \right) = RHS
    $ and finally : $\displaystyle
    \log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1}
    {n}} \right) \leq \log \left( n \right) - 1 + \tfrac{1}
    {{2n}} < \log \left( n \right) - 1 + \tfrac{1}
    {n} \leq \tfrac{1}
    {n} \cdot \log \left( {n!} \right)
    $


    In fact, the LHS of your inequality ( as written originally) converges to 'e'.
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  3. #3
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    It's seem be ok,only i have one question:
    WHY
    $\displaystyle
    \int_0^n {\tfrac{{\left\{ x \right\}}}
    {x}dx} \geqslant \int_0^1 {\tfrac{{\left\{ x \right\}}}
    {x}dx} = \int_0^1 {dx}
    $
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  4. #4
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
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    571
    Because $\displaystyle \tfrac{{\left\{ x \right\}}} {x}\geq 0$ when $\displaystyle x\in (0,+\infty)$ implies that the sequence $\displaystyle a_n=\int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx}$ is non-decreasing ( $\displaystyle n\in {\mathbb{Z}^+}$ ) and so $\displaystyle a_n\geq{a_1}$.

    Now when $\displaystyle x\in (0, 1)$ we have $\displaystyle {\left\{ x \right\}}=x$, and so $\displaystyle \int_0^1 {\tfrac{{\left\{ x \right\}}} {x}dx} = \int_0^1 {dx}=1$
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