<s>[INST] Proving that: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$ Let $a$ and $b$ be positive reals. Show that
$$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$
 [/INST] 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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