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Math Help - Limit of a sequence - nice one

  1. #1
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    Limit of a sequence - nice one

    Well this one looks very interesting, it looks pretty obvious but I can't find a way to prove it formally:

    Let a,b,c be positive numbers.
    Prove:
    lim(a^n+b^n+c^n)^\frac{1}{n} = max(a,b,c)
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  2. #2
    Super Member Bacterius's Avatar
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    Try to define mathematically what max(a, b, c) is (I believe it is the greatest of these numbers ?). You can't just leave a verbal operation in your limit, you must express it in algebra´c operations. It might be easier
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  3. #3
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    Quote Originally Posted by adam63 View Post
    Let a,b,c be positive numbers.
    Prove: lim(a^n+b^n+c^n)^\frac{1}{n} = max(a,b,c)
    Here is the basic idea.
    \max \left\{ {a,b,c} \right\} = M = \sqrt[n]{{M^n }} \leqslant \sqrt[n]{{a^n  + b^n  + c^n }} \leqslant M\sqrt[n]{3}
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    MHF Contributor chisigma's Avatar
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    Let's suppose that a>b and a>c [othervise you can always swap the variables ... ]. Because is...

    \sqrt[n]{a^{n} + b^{n} + c^{n}} = a \sqrt{1+(\frac{b}{a})^{n} + (\frac{c}{a})^{n}} (1)

    ... and...

    \frac{b}{a}<1 , \frac{c}{a}<1 (2)

    ... from (1) and (2) we derive...

    \lim_{n \rightarrow \infty} \sqrt[n]{a^{n} + b^{n} + c^{n}} = a (3)

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Let's suppose that a>b and a>c [othervise you can always swap the variables ... ]. Because is...

    \sqrt[n]{a^{n} + b^{n} + c^{n}} = a \sqrt{1+(\frac{b}{a})^{n} + (\frac{c}{a})^{n}} (1)

    ... and...

    \frac{b}{a}<1 , \frac{c}{a}<1 (2)

    ... from (1) and (2) we derive...

    \lim_{n \rightarrow \infty} \sqrt[n]{a^{n} + b^{n} + c^{n}} = a (3)

    Kind regards

    \chi \sigma
    very nice !!
    I like such proofs!
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