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Math Help - Inequality

  1. #1
    Newbie
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    Inequality

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

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

    Hence we have: LHS=<br />
\log \left( n \right) - n \cdot \log \left( {1 + \tfrac{1}<br />
{n}} \right) \leq \log \left( n \right) - 1 + \tfrac{1}<br />
{{2n}}<br />

    Next note that: <br />
\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\}}}<br />
{x}dx}<br />
- see Lemma 1, here-

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

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


    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
    <br />
\int_0^n {\tfrac{{\left\{ x \right\}}}<br />
{x}dx} \geqslant \int_0^1 {\tfrac{{\left\{ x \right\}}}<br />
{x}dx} = \int_0^1 {dx} <br />
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  4. #4
    Super Member PaulRS's Avatar
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    Because \tfrac{{\left\{ x \right\}}} {x}\geq 0 when x\in (0,+\infty) implies that the sequence a_n=\int_0^n {\tfrac{{\left\{ x \right\}}} {x}dx} is non-decreasing ( n\in {\mathbb{Z}^+} ) and so a_n\geq{a_1}.

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