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If $a_1,a_2,\dotsc,a_n $ are positive real numbers, then prove that

$$\lim_{x \to \infty} \left[\frac {a_1^{1/x}+a_2^{1/x}+.....+a_n^{1/x}}{n}\right]^{nx}=a_1 a_2 \dotsb a_n.$$

My Attempt:

Let $P=\lim_{x \to \infty} \left[\dfrac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} \implies \ln P=\lim_{x \to \infty} \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} =\lim_{x \to \infty} nx \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]= \lim_{x \to \infty} n \left[\frac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\right]$

and this is $0/0$ form and so I have to apply L'Hospital's rule. Now things get a bit complicated during derivative.

Can someone point me in the right direction? Thanks in advance for your time.

BLAZE
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learner
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  • First perform the substitution $y=\frac{1}{x}$ so that you are taking the limit as $y \to 0$. Now use L'Hopital's rule. – Ivan Loh Apr 10 '13 at 14:02
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    I should note that your attempt contains a bit of poor maths grammar. "Let $P = S \implies \ldots$ If you try reading this out loud, you'd get "If Let $P = S$, then $\ldots$" or "Let $P=S$ implies $\ldots$" neither of which make a whole lot of sense. Better would be the following: "Let $P = S$. Then, applying the natural logarithm to both sides, we get that $\ldots$" – kahen May 28 '13 at 05:50

3 Answers3

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Let $$\begin{align}P &=\lim_{x \to \infty} \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]^{nx}\\ \implies \ln P &=\lim_{x \to \infty} \ln \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]^{nx} \\&=\lim_{x \to \infty} nx \ln \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]\\&= \lim_{x \to \infty} n \Big[\dfrac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\Big]\\&=\lim_{z \to 0} n \Big[\dfrac {\ln (a_1^{z}+a_2^{z}+...+a_n^{z})-\ln n}{z}\Big]\end{align}$$

and this is $0/0$ form and so I have to apply L'Hospital's rule. So,$$\begin{align}\ln P &=n \lim_{z \to 0}\dfrac {1}{ (a_1^{z}+a_2^{z}+...+a_n^{z})} \times \{a_1^z \ln a_1+a_2^z \ln a_2+.....+a_n^z \ln a_n\} \\&=n \times \dfrac {1}{n}\{\ln a_1+\ln a_2+....+\ln a_n\}\\&=\ln (a_1.a_2...a_n)\\\implies P&=a_1.a_2...a_n \end{align}$$.

This completes the proof.

learner
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  • I have used @Ivan Loh's suggestion for solving it. Did I miss anything? – learner Apr 10 '13 at 14:25
  • Once you have proved that the limit P does indeed exist, then yes. Personally, I would also use that ln is continuous for all x > 0 as an explanation for why the top line holds, but the general argument does seem fine. – Andrew D Apr 10 '13 at 15:15
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I will just prove this limit $$ \lim_{x\to\infty} \left(\frac {a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.....+a_n^{\frac{1}{x}}}{n}\right)^{x}=(a_1.a_2.....a_n)^{\frac{1}{n}} $$ We have $$\begin{align} \lim_{x\to\infty} \left(\frac {a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.....+a_n^{\frac{1}{x}}}{n}\right)^{x}&=\lim_{x\to\infty} \left(1+\frac {a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.....+a_n^{\frac{1}{x}}}{n}-1\right)^{x} \\&=e^{\displaystyle\lim_{x\to\infty}x\left(\dfrac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right)} \end{align}$$ we know $a^x-1$ ~ $x\ln a$

so $$ e^{\displaystyle\lim_{x\to\infty}\left(x\cdot\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right)}=e^{\dfrac{(\ln a_1+\ln a_2+...+\ln a_n)}{n}}=(a_1a_2\cdots a_n)^{\frac{1}{n}} $$

antshar
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Xiaolang
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Another method to resolve this problem is, take the inequality $GM \leqslant AM \leqslant x \text {th root mean}$, then use squeeze theorem.