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I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:

$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$

As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:

$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$

But now when I approach that top limit this is what I get:

$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$

And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?

Grigory M
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F.Webber
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4 Answers4

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By Taylor series we have: $$\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}=\frac 1 3\left(3+\frac1 n(\log a +\log b+\log c)++o\left(\frac 1 n\right)\right)=1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)$$ so $$\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\exp\left(n\log\left(1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)\right)\right)\sim_\infty \sqrt[3]{abc} $$

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While $\sqrt[n]a\to 1$, it is not correct to say $(n\sqrt[n]a-n)\to 0$. Actually, $\sqrt[n]{1+\epsilon}\approx 1+\frac1n\epsilon$ so $n\sqrt[n]a-n\approx a$.

Hagen von Eitzen
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  • Oh thats a pretty neat approximation, I'll keep that in mind for the future, though is that approximation easy to obtain? – F.Webber Jan 09 '14 at 19:04
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This is not an answer to the question itself I think, but still a useful observation.

This can be found using the Generalized means. Say that $$ M_p(x_1,\dots,x_n)=\left(\frac 1n\sum x_i^p\right)^\frac 1p $$ In general, we have $M_m(x_1,\dots,x_n)\geq M_n(x_1,\dots,x_n)$ when $m\geq n$, with equality if and only if $x_i=x_j$ for all $i$ and $j$. Also, (I now realize THIS is your question, although I'll just assume it's true), we have $$M_0(x_1,\dots,x_n)=\sqrt[n]{x_1\cdots x_n}$$ Thus, we have $$\lim_{p\to 0} M_p(a,b,c)=M_0(a,b,c)=\sqrt[3]{abc}$$

Ragnar
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Fact I. $$ \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n^2}\right)^{\!n}=\mathrm{e}^a. $$

Fact II. For every $a>0$, there exists a $b>0$, such that $$ 1+\frac{\ln a}{n}\le a^{1/n} \le 1+\frac{\ln a}{n}+\frac{b}{n^2}. $$

Using the two facts: $$ 1+\frac{\ln a+\ln b+\ln c}{3n}\le\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\le 1+\frac{\ln a+\ln b+\ln c}{3n}+\frac{k}{n^2} $$ and hence $$ \left(1+\frac{\ln\sqrt[3]{abc}}{n}\right)^{\!n}\le \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{\!n} \le\left(1+\frac{\ln(\sqrt[3]{abc})}{n}+\frac{k}{n}\right)^{\!n} $$ and as the both the left and right hand side tend to $\sqrt[3]{abc}$, so does the middle one.

Yiorgos S. Smyrlis
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