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For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$

how can I find the limit of sequence above? Provide me a hint or full solution. thanks ^^

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  • $\begingroup$ Suppose (without loss of generality) that $a\le b \le c$. Then $\lim {\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}3\right)}^n \le \lim {\left(\frac{c^{1/n}+c^{1/n}+c^{1/n}}3\right)}^n = c$… $\endgroup$ – MJD May 20 '13 at 16:49
  • $\begingroup$ Again, $$\frac{a^\frac1n+b^\frac1n+c^\frac1n}3\ge (abc)^{\frac1{3n}}\implies \left(\frac{a^\frac1n+b^\frac1n+c^\frac1n}3\right)^n\ge \sqrt[3]{abc} $$ $\endgroup$ – lab bhattacharjee May 20 '13 at 16:51
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$$\lim_{n \to \infty}\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n=\lim_{n \to \infty}\frac{\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)}{\frac{1}{n}}=\lim_{x\to 0^+}\frac{\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)}{x}$$

L'Hospital or the definition of the derivative solves it. Actually since the limit is just the definition of the derivative of $\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)$ at $x=0$, it would be wrong to use L'H :)

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This answer is inspired by @Aryabhata's answer here.

Suppose $x_k>0$. By $\mathrm{AM}\ge\mathrm{GM}$ we have $$\sqrt[m]{x_1\cdots x_m} \le \frac{x_1+\ldots +x_m}{m}.$$ By weighted $\mathrm{AM}\ge\mathrm{GM}$ we have $$\begin{eqnarray*} \frac{m}{x_1+\ldots +x_m} &=& \frac{x_1\cdot 1/x_1+\ldots +x_m\cdot 1/x_m}{x_1+\ldots +x_m} \\ %&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{1/(x_1+\ldots +x_m)} \\ %&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{\frac{1}{x_1+\ldots +x_m}} \\ &\ge& \sqrt[x_1+\ldots +x_m]{\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}}. \end{eqnarray*}$$ Therefore, $$\sqrt[m]{x_1\cdots x_m} \le \frac{x_1+\ldots +x_m}{m} \le \sqrt[x_1+\ldots +x_m]{x_1^{x_1}\cdots x_m^{x_m}}.$$

Letting $x_k = a_k^{1/n}$ and taking the $n$th power we find by the squeeze theorem that $$\lim_{n \to \infty}\left(\frac{a_1^{1/n}+\ldots +a_m^{1/n}}{m}\right)^n = \sqrt[m]{a_1\cdots a_m}.$$ In particular, $$\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n = \sqrt[3]{a b c}.$$

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$$\begin{align}\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{n} &=\lim_{r \to 0^+}\left(\frac{a^{r}+b^{r}+c^{r}}{3}\right)^{1/r}\\ &=\lim_{r \to 0^+}\frac{1}{r}\cdot \exp\left( [\log \left(a^{r}+b^{r}+c^{r}\right)]\right) \\ &=\exp\left( [\lim_{r \to 0}\frac{d}{dr}\log(a^{r}+b^{r}+c^{r})]\right) \\ &=\exp\left( \frac{[\log(a)+\log(b)+\log(c)]}{3}\right) \\ &= (abc)^{1/3} \end{align}$$

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  • $\begingroup$ I have edited your answer as it was quite startling to see such huge fonts. If you want to see the $\TeX$ I used, you can right click on the math and select "Show Math As -> TeX Commands" $\endgroup$ – JavaMan May 21 '13 at 4:50
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$a^{\frac 1n}=1 +\dfrac{\log a}{n}+O(\frac1{n^2})$

So,$\Big(\dfrac{a^{\frac 1 n}+b^{\frac 1 n}+c^{\frac 1 n}}3\Big)^n=(1+\frac{\log abc}{3n}+O(\frac1{n^2}))^n\to \sqrt[3]{abc}$ This generalizes easily to 4 or more terms also.

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