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

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

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


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

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


    Since this simplifies to $\displaystyle \displaystyle \frac{0}{0}$, 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 $\displaystyle a_n=f(n)$ from $\displaystyle \mathbb{N}$ to $\displaystyle \mathbb{R}^+-\{0\}$ the limit is:

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

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

    $\displaystyle L=\sqrt[3]{abc}$


    Fernando Revilla
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