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.
 A: Taking the "redefinition" as $\prod_{i=1}^\infty a^{\frac 1{3^i}}$, we immediately have $\sum_{i=1}^\infty \frac 1{3^i}=\frac 12$, which is the effect of summing the exponents as would occur with the product described.  Therefore, we have 
$$\prod_{i=1}^\infty a^{\frac 1{3^i}}=\sqrt a$$
A: Another way to go is to show that $x_n=a^{b_n}$ where $b_n$ satisfies
$$b_1=\frac13\quad\text{and}\\
b_{n+1}=(1+b_n)\frac13$$
then show $\lim b_n=\frac12$ in a manner similar to user17762's answer.
A: Let $x_n = \sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\cdots}}}}}}}}}$, where the $\sqrt[k]{\cdot}$ appears $n$ times. Prove that this sequence is monotone and bounded. Now Setting $\lim_{n \to \infty} x_n = x$, we obtain
$$x = \sqrt[k]{ax} \implies x^k = ax \implies x^{k-1} = a \implies x = \sqrt[k-1]{a}$$
In your case, $k=3$.
A: Let $\displaystyle b=\sqrt[3]{a\cdot\  \underbrace{\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}}}}}}}}}_{\text{This is $b$, which is allegedly $\sqrt a$.}}}$.
Then $\displaystyle b = \sqrt[3]{ab}$, so $b^3 = ab$, and then $b^2 = a$.  Thus $b = \sqrt a$.
If we put $\sqrt{a}$ in place of the expression said to be $b=\sqrt{a}$, we get $\sqrt[3]{a\sqrt{a}}$.  And it is easy to see that that is indeed $\sqrt a$.
As to convergence: let $g(x) = (ax)^{1/3}$. The question is the behavior of the sequence
$$
a,\ g(a),\ g(g(a)),\  g(g(g(a))),\ \ldots\ .
$$
For $x$ between $\sqrt{a}$ and $a$, we have $0<g'(x)<1/3$, so $g$ is a contraction and thus has a unique attractive fixed point.
A: 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 $a^{\frac{1}{2}}=\sqrt{a}$
A: this might not be entirely rigorous, but here is a simple intuitive way to think of it: \begin{align*}x &= \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}}}}}}}}}\\
x^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\sqrt[3]{a\cdots}}}}}}}}}\\
\frac{x^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\sqrt[3]{a\cdots}}}}}}}}}\\
\frac{x^3}{a} &= x\\
x^2&=a\\
x&=\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}
\end{align*}
