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$$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}=\sqrt{a}$$

Just for fun. I would like to read the proof of this if it exists. Any references would be appreciated.

Martin Sleziak
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Ivan
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    As a first step, try to define the sequence rigorously, i.e., give the explicit recursive definition. Then you should be able to show the sequence is monotonic, bounded, and then figure out which equation does the limit satisfy. – Ittay Weiss Apr 09 '15 at 23:41
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    So this is equivalent to $$\prod_{i=1}^\infty a^{\frac 1{3^i}}$$ – abiessu Apr 09 '15 at 23:42
  • @abiessu hahah oh wow I did not see that. It's easy then. Thank you! – Ivan Apr 09 '15 at 23:53
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    Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Apr 10 '15 at 11:09
  • This is almost the same question as http://math.stackexchange.com/questions/815418/why-does-sqrtn-sqrtn-sqrtn-ldots-n – DanielV Jun 03 '15 at 01:56

6 Answers6

16

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, $k=3$.

Michael Hardy
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Adhvaitha
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    $$x_{n+1}=\sqrt[k]{ax_n},$$ $$x=\lim_{n\to \infty} x_n= \lim_{n\to \infty} x_{n+1} =\lim_{n\to \infty} \sqrt[k]{ax_n}=\sqrt[k]{a \lim_{n\to \infty} x_n}=\sqrt[k]{ax}$$ – Leonhardt von M Apr 10 '15 at 00:00
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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.

Michael Hardy
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6

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 $$\prod_{i=1}^\infty a^{\frac 1{3^i}}=\sqrt a$$

abiessu
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4

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.

Tim Raczkowski
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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 $a^{\frac{1}{2}}=\sqrt{a}$

Ahmed S. Attaalla
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    You should write your answer in $\rm\LaTeX$. I think you might have the right idea more or less and is quite similar in nature to @abiessu's answer. However it's hard to tell when it's written like this. – Cameron Williams Jun 03 '15 at 00:30
  • May I ask you to join a chat room of mine? The chat room is at the bottom of my profile description. – Simply Beautiful Art Feb 02 '17 at 01:27
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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*}

nosyarg
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