# Provide different proofs for the following equality: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$

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

#### Solutions Collecting From Web of "Provide different proofs for the following equality: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$"

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,
$$\lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n = \sqrt{a b}.$$

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.

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,
$$\begin{eqnarray*} \lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=& \lim_{n\to\infty} b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n \\ &=& b e^{\ln \sqrt{\frac{a}{b}}} \\ &=& \sqrt{a b}. \end{eqnarray*}$$

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}

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.

Let us first consider that case that $ab=1$, i.e., $b=1/a$. So we want to show find the limit
$$\lim\limits_{n\to\infty} \left(\frac {a^{1/n}+\frac1{a^{1/n}}}2\right)^n = 1. \tag{1}$$
W.l.o.g. we may assume that $a\ge1$. (Otherwise we replace $a$ by $1/a$.)

We can use the fact that for $x\ge1$ we have $2\le x+\frac1x \le 2+(x-1)^2$, which can be verified by a straightforward algebraic manipulation.
(Just notice that $x+\frac1x-2 = \frac{x^2-2x+1}x = \frac{(x-1)^2}x$ and $2+(x-1)^2-x-\frac1x= \frac{x^3-3x^3+3x-1}x = \frac{(x-1)^3}x$.)

So we have
$$1 \le \frac{x+\frac1x}2 \le 1+\frac12(x-1)^2\tag{2}$$
for $x\ge1$.

Using $(2)$ for $x=a^{1/n}$ we get
$$1 \le \left(\frac {a^{1/n}+\frac1{a^{1/n}}}2\right)^n \le \left(1+\frac12(a^{1/n}-1)^2\right)^n \le 1+\frac n2 (a^{1/n}-1)^2.\tag{3}$$
Now it suffices to notice that
$$\lim\limits_{n\to\infty} \frac n2 \left(a^{1/n}-1\right)^2 = \lim\limits_{n\to\infty} \frac 1{2n} \left(\frac{a^{1/n}-1}{1/n}\right)^2 = 0.$$
(Here we are using that $\lim\limits_{x\to0} \frac{a^x-1}x = \ln a$, which implies that $(\frac{a^{1/n}-1}{1/n})^2$ is bounded.)

So by squeeze theorem we get that all expressions in $(3)$ converge to $1$ for $n\to\infty$, which proves $(1)$.

Once we have $(1)$ we can prove general case using
$$\left(\frac{a^{1/n}+b^{1/n}}2\right)^n = \sqrt{ab} \left(\frac{\left(\sqrt{\frac{a}{b}}\right)^{1/n}+\left(\sqrt{\frac{b}{a}}\right)^{1/n}}2\right)^n.$$
We get the same limit as in $(1)$ with $\sqrt{\frac{a}{b}}$ instead of $a$.

The proof is based on the use of the standard limits $$\lim_{n \to \infty}n(a^{1/n} – 1) = \log a\tag{1}$$ (which can also be taken as a definition of $\log a$ to develop a systematic theory of logarithm and exponential functions) and $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{2}$$ If $L$ is the desired limit then we have
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{a^{1/n} + b^{1/n}}{2}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{a^{1/n} + b^{1/n}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\log\left(\frac{a^{1/n} + b^{1/n}}{2}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{a^{1/n} + b^{1/n} – 2}{2}\cdot\dfrac{\log\left(1 + \dfrac{a^{1/n} + b^{1/n} – 2}{2}\right)}{\dfrac{a^{1/n} + b^{1/n} – 2}{2}}\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{a^{1/n} + b^{1/n} – 2}{2}\cdot 1\text{ (using (2))}\notag\\
&= \frac{1}{2}\lim_{n \to \infty}\left\{n(a^{1/n} – 1) + n(b^{1/n} – 1)\right\}\notag\\
&= \frac{1}{2}(\log a + \log b)\text{ (using (1))}\notag\\
&= \log\sqrt{ab}\notag
\end{align}
and hence $L = \sqrt{ab}$. Same way we can establish the general formula that if $a_{i}$ are positive then $$\lim_{n \to \infty}\left(\frac{a_{1}^{1/n} + a_{2}^{1/n} + \cdots + a_{m}^{1/n}}{m}\right)^{n} = (a_{1}a_{2}\cdots a_{m})^{1/m}\tag{3}$$

Let $f(x) = \ln [(a^x+b^x)/2].$ Note that $f$ is differentiable everywhere, and $f(0)=0.$ As $x\to 0,$

$$\tag 1\frac{\ln [(a^x+b^x)/2]}{x}=\frac{f(x) – f(0)}{x-0} \to f'(0).$$

Computing $f'(0)$ is straightforward: We get $f'(0)= \ln (ab)^{1/2}.$
Taking $x= 1/n$ and using $(1)$ then gives

$$\frac{\ln [(a^{1/n}+b^{1/n})/2]}{1/n} = \ln [(a^{1/n}+b^{1/n})/2)]^n\to \ln (ab)^{1/2}.$$ Exponentiating gives $[(a^{1/n}+b^{1/n})/2]^n \to (ab)^{1/2}$ as desired.