Provide different proofs for the following equality.

Let $a$ and $b$ be positive reals. Show that $$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$

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There are infinitely many possible proofs. – Bill Cook Apr 11 '12 at 17:01
Even then you should clarify when two solutions are assumed to be "different" to you. Multiplying by some number, doing a simplification and then divising that number again probably doesn't count as "different"? ;) – example Apr 11 '12 at 17:10
The collection of possible proofs of a statement isn't a set, and it shouldn't be thought of as having a cardinality. – Qiaochu Yuan Apr 11 '12 at 17:11
Take a standard proof of this statement. Pick a random set $S$ and insert (the irrelevant) statement: "Consider $x \in S$." Since the set of all sets is not a set. This provides a different proof for each set $S$ and thus the collection of all possible proofs is not a set. – Bill Cook Apr 11 '12 at 17:17
However, if you are working within a fixed proof system with a fixed set of accepted symbols (finitely many). Then there are only countably many finite strings over a finite alphabet so there are only countably many proofs within such a fixed system. – Bill Cook Apr 11 '12 at 17:20

Another proof. Expand in a binomial series, $$\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n = \frac{1}{2^n} \sum_{k=0}^n {n\choose k} (a^{1/n})^k (b^{1/n})^{n-k}.$$ Use the de Moivre-Laplace theorem, $${n\choose k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{n-k} (a^{1/n})^k (b^{1/n})^{n-k} \simeq \frac{1}{\sqrt{2\pi}\sigma} e^{-(k-\mu)^2/(2\sigma^2)} (a^{1/(2\mu)})^k (b^{1/(2\mu)})^{2\mu - k}$$ where $\mu = n/2$ and $\sigma^2 = n/4$. Change variables. Let $z = (k-\mu)/\sigma$. Therefore, $$\lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n = \lim_{n\to\infty} \sqrt{a b} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dz \, e^{-z^2/2} \left(\frac{a}{b}\right)^{\sigma z/(2\mu)}.$$ The integral can be done easily enough by completing the square. We find $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dz \, e^{-z^2/2} \left(\frac{a}{b}\right)^{\sigma z/(2\mu)} = \exp \frac{\sigma^2 \log^2(a/b)}{8\mu^2}.$$ But $\sigma/\mu = 1/\sqrt{n}$. Therefore, in the limit the integral is unity. Thus, $$\lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n = \sqrt{a b}.$$

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Thanks. Indeed, I wanted this approach for my problem. Thanks for noting me that. :) – Babak S. Dec 23 '12 at 19:05
@BabakSorouh: Glad to help, Babak. Cheers! – oen Dec 23 '12 at 19:45

You can use the following inequality:

$$\sqrt{xy} \le \frac{x+y}{2} \le \sqrt[x+y]{x^x y^y}$$

The first inequality is straightforward, and the second one can be gotten by

$$\frac{2}{x+y} = \frac{ x \times 1/x + y \times 1/y}{x+y} \ge \sqrt[x+y]{\frac{1}{x^x y^y}}$$

using the weighted $\text{AM} \ge \text{GM}$.

Setting $x = a^{1/n}$, $y = b^{1/n}$ and taking the $n^{th}$ powers gives us that the limit is $\sqrt{ab}$, by the squeeze theorem.

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NOTE: Given the OP asks for a collection of proofs, I guess it is appropriate to make this a wiki post.

You want to prove

$$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{a^{1/n}} + {b^{1/n}}}}{2}} \right)^n} = \sqrt {ab}$$

Assume that $a <b$, since $a=b$ will trivially yield the result. We have an indeterminate for of $1^\infty$.

We use

$$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{a^{1/n}} + {b^{1/n}}}}{2}} \right)^n} = \exp \mathop {\lim }\limits_{n \to \infty } n\log \left( {\frac{{{a^{1/n}} + {b^{1/n}}}}{2}} \right)$$

Then we reduce the indetermination to one of the form $\infty \cdot0$ which is then reduced to one of the form $0/0$, namely:

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\log \left( {\frac{{{a^{1/n}} + {b^{1/n}}}}{2}} \right)}}{{\frac{1}{n}}}$$

Given no assumption is made on $n$ I use L'Hôpital's Rule, from where

$$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{a^{1/n}} + {b^{1/n}}}}{2}} \right)^n} = \exp \mathop {\lim }\limits_{n \to \infty } \frac{{{a^{1/n}}\log a + {b^{1/n}}\log b}}{{ - 2{n^2}}}\frac{{ - {n^2}}}{{\frac{{{a^{1/n}} + {b^{1/n}}}}{2}}}$$

$$= \exp \mathop {\lim }\limits_{n \to \infty } \frac{{{a^{1/n}}\log a + {b^{1/n}}\log b}}{{{a^{1/n}} + {b^{1/n}}}}$$

Now this yields

$$\exp \frac{{\log a + \log b}}{2} = \exp \log \sqrt {ab} = \sqrt {ab}$$

If $n$ is a discrete variable you can use L'Hôpital's discrete analog.

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An elementary proof. We use the Taylor series $e^x = 1 + x + O(x^2)$ and the fact that $\lim_{n\to\infty}(1+x/n)^n = e^x$.

If $a=b$ the identity is trivial. Without loss of generality, assume $0<a<b$. Then $$\begin{eqnarray*} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=& b \left(\frac{1+(\frac{a}{b})^{1/n}}{2}\right)^n \\ &=& b \left(\frac{1+e^{\frac{1}{n}\ln \frac{a}{b}}}{2}\right)^n \\ &=& b \left(1+\frac{1}{2}\frac{1}{n}\ln \frac{a}{b} + O(1/n^2)\right)^n \\ &=& b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n + O(1/n). \end{eqnarray*}$$ Therefore, $$\begin{eqnarray*} \lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=& \lim_{n\to\infty} b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n \\ &=& b e^{\ln \sqrt{\frac{a}{b}}} \\ &=& \sqrt{a b}. \end{eqnarray*}$$

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Why do you call this elementary? – Peter Tamaroff Apr 12 '12 at 1:30
@PeterT.off: In what sense is it not? – oen Apr 12 '12 at 1:39
Taylor series, $o$ notation. I'd say what is elementary is Aryabhata's proof. – Peter Tamaroff Apr 12 '12 at 1:43
@PeterT.off: It is fairly typical to label a proof only depending on real analysis "elementary." Notice that I did not claim that the other proofs were not elementary. – oen Apr 12 '12 at 1:58
This was the first way I thought of going. (+1) – robjohn Apr 12 '12 at 3:19
For large $n$, $$x^{1/n}=1+\frac1n\log(x)+O\left(\frac{1}{n^2}\right)$$ Thus, \begin{align} \lim_{n\to\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=\lim_{n\to\infty}\left(1+\frac1n\left(\frac{\log(a)+\log(b)}{2}\right)+O\left(\frac{1}{n^2}\right)\right)^n\\ &=\exp\left(\frac{\log(a)+\log(b)}{2}\right)\\ &=\sqrt{ab} \end{align}