<s>[INST] Can this be shown: $\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\dots}}} = \sqrt a$? $$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}=\sqrt{a}$$
Just for fun. I would like to read the proof of this if it exists. Any references would be appreciated.
 [/INST] What we have is the $lim_{n\to\infty}$ $a^{\frac{1}{3}}a^{\frac{1}{9}}a^{\frac{1}{27}}..a^{\frac{1}{3^n}}$.
Using the property $(a^b)(a^c)=a^(b+c)$
We get:
$a^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...\frac{1}{3^n}}$
The exponent is in fact is geometric series which has a value of $(1/(1-(1/3))-1=\frac{1}{2}$ so:
We get your expression to be $a^{\frac{1}{2}}=\sqrt{a}$
</s>