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Math Help - lim (n-->infinity){ [a^(1/n)+b^(1/n)+c^(1/n)]/3}^n

  1. #1
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    lim (n-->infinity){ [a^(1/n)+b^(1/n)+c^(1/n)]/3}^n

    For a, b, c > 0

    What is lim (n-->infinity){ [a^(1/n)+b^(1/n)+c^(1/n)]/3}^n equal to ?

    The limit converges, as I have tried to similuate the number. The answer must be in terms of a, b, and c.

    I tried to expand by binomial series, but got to nowhere useful.

    If anyone could help, I would be appreciated. Thank you very much.
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  2. #2
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    \displaystyle \left(\frac{a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}}{3}\right)^n = e^{\ln{\left[\left(\frac{a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}}{3}\right)^n\right]}}

    \displaystyle = e^{n\ln{\left(\frac{a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}}{3}\right)}}

    \displaystyle = e^{\frac{\ln{\left(a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}\right)} - \ln{3}}{\frac{1}{n}}}


    So \displaystyle \lim_{n \to \infty}\left(\frac{a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}}{3}\right)^n = \lim_{n \to \infty}e^{\frac{\ln{\left(a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}\right)} - \ln{3}}{\frac{1}{n}}}

    \displaystyle = e^{\lim_{n \to \infty}\frac{\ln{\left(a^{\frac{1}{n}} + b^{\frac{1}{n}} + c^{\frac{1}{n}}\right)} - \ln{3}}{\frac{1}{n}}}.


    Since this simplifies to \displaystyle \frac{0}{0}, you can use L'Hospital's Rule to evaluate the limit.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    An alternative. Expanding in a natural way the function a_n=f(n) from \mathbb{N} to \mathbb{R}^+-\{0\} the limit is:

    L=e^{\lambda},\quad \lambda=\displaystyle\lim_{x \to{+}\infty}{x\left(\frac{a^{1/x}+b^{1/x}+c^{1/x}}{3}-1\right)}

    Now, use the substitution t=1/x and the L'Hopital's rule. You'll obtain:

    L=\sqrt[3]{abc}


    Fernando Revilla
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