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How would one compute $$\lim_{n\to\infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}3\right)^n$$ if $a,b,c>0$.

I've never done an integral 3 variables before! This is from a chapter called Interchange of Limit Operations. Ratio test is fair game but I don't know if that will help.

Stephanie Wayne
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2 Answers2

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We remark that

$$ \lim_{n\to\infty} \left\{ 1 + \frac{x}{n} + o\left(\frac{1}{n}\right) \right\}^{n} = e^{x}. $$

In this case, we have

$$ \frac{a^{1/n} + b^{1/n} + c^{1/n}}{3} = 1 + \frac{\log(abc)}{3n} + o\left(\frac{1}{n}\right) $$

and we have the answer

$$ \lim_{n\to\infty} \left( \frac{a^{1/n} + b^{1/n} + c^{1/n}}{3} \right)^{1/n} = \sqrt[3]{abc}. $$

Sangchul Lee
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  • Hmm, that looks right, but then my answer is wrong. Not sure where my answer is wrong. – Thomas Andrews Sep 25 '13 at 19:43
  • @ThomasAndrews, the error term cannot be of order $1/n^{2}$ since $$ \binom{1/n}{k} \sim \frac{(-1)^{k-1}}{nk} \quad \text{as} \quad n \to \infty $$ for each $k = 1, 2, \cdots$. – Sangchul Lee Sep 25 '13 at 19:54
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Another approach:

$$a_n:=\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}3\right)^n\implies \log a_n=n\log\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}3\right)$$

But applying l'Hospital's rule to the continuous variable function we get

$$\lim_{n\to\infty}\log a_n=\lim_{n\to\infty}\frac{\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}3\right)}{\frac1n}\stackrel{\text{l'H}}=\lim_{n\to\infty}\frac{-\frac1{3n^2}\left(a^{1/n}\log a+b^{1/n}\log b+c^{1/n}\log c\right)}{-\frac1{n^2}}=$$

$$=\frac13\left(\log a+\log b+\log c\right)=\log \sqrt[3]a+\log\sqrt[3]b+\log\sqrt[3]c\implies$$

$$\implies\lim_{n\to\infty}a_n=\sqrt[3]a\sqrt[3]b\sqrt[3]c=\sqrt[3]{abc}$$

DonAntonio
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