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Thread: 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:
    $\displaystyle lim(a^n+b^n+c^n)^\frac{1}{n} = max(a,b,c)$
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  2. #2
    Super Member Bacterius's Avatar
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    Try to define mathematically what $\displaystyle max(a, b, c)$ 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: $\displaystyle lim(a^n+b^n+c^n)^\frac{1}{n} = max(a,b,c)$
    Here is the basic idea.
    $\displaystyle \max \left\{ {a,b,c} \right\} = M = \sqrt[n]{{M^n }} \leqslant \sqrt[n]{{a^n + b^n + c^n }} \leqslant M\sqrt[n]{3}$
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    MHF Contributor chisigma's Avatar
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    Let's suppose that $\displaystyle a>b$ and $\displaystyle a>c$ [othervise you can always swap the variables ... ]. Because is...

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

    ... and...

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

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

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

    Kind regards

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

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

    ... and...

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

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

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

    Kind regards

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