1. ## 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.

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

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

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

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.

3. 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