<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] 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.
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