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Math Help - [SOLVED] inequalities related to binomial theorem

  1. #1
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    [SOLVED] inequalities related to binomial theorem

    (1+\frac{1}{n})^n\ge2, \forall n \in \mathbb{Z} : n>0

    So the easiest way I think is to use binomial theorem.

    I've gotten to:

    \Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}+1

    So I guess I essentially have to prove:

    \Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}>1
    Last edited by seld; September 9th 2009 at 07:41 PM. Reason: forgot to add the n in N statement
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  2. #2
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    Quote Originally Posted by seld View Post
    (1+\frac{1}{n})^n\ge2

    So the easiest way I think is to use binomial theorem.

    I've gotten to:

    \Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}+1

    So I guess I essentially have to prove:

    \Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}>1
    There is an alternative approach that is not hard.

    Let f(x) = \left(1+\frac{1}{x}\right)^x - 2 for x>0.
    Show that f ' (x) > 0 therefore f is increasing.
    At x=2 you can easily calculate that f(2) > 0.
    Therefore, if x\geq 2 then since it is increasing it means f(x)\geq f(2) > 0.
    But if f(x) > 0 \implies \left(1+\tfrac{1}{x}\right)^x > 2.
    Therefore, this inequality is true for any n\geq 2.
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  3. #3
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    nice, though i'm not allowed to use derivatives and such yet, we're only allowed to use inequality propositions, order axioms for R, field axioms, absolute value propositons. I'm basically taking a proofs class, and the book even says I'm supposed to use binomial theorem.

    and I guess I forgot to mention it's \forall n \in \mathbb{Z} : n > 0.

    I had thought about just doing:

    (1+\frac{1}{n})^n=(\frac{n+1}{n})^n\ge2

    then doing induction on the statement though I haven't made much headway.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Multiplying through by n^n, it's equivalent to showing

    (1+n)^n>2n^n

    To do that just note that

    \sum_{j=0}^n{n \choose j}n^j\geq 1 + {n \choose n-1}n^{n-1}+n^n
     =1+n\times n^{n-1}+n^n = 1+2n^n


    That's it

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  5. #5
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    thanks, though i feel stupid now, because that's exactly the same thing I used to prove something earlier.
    Last edited by seld; September 9th 2009 at 09:13 PM. Reason: misread solution
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    I haven't assumed the result is true. What I have done is I've shown that (1+n)^n>2n^n (you agree?). Now just divide this inequality by n^n and you have the result.
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