# Infinite radical. How to show it converges to three?

Recently i have discovered a new infinite radical with non-constant (that means they are not the same and not periodic) coefficients. But i don't have a strict proof of this identity... The identity is

$$\sqrt[3]{23+\sqrt[3]{54+\sqrt[3]{972+\sqrt[3]{21870+\sqrt[3]{551124+\sqrt[3]{14526054+...}}}}}}=3.$$

The numbers are (besides 23) $3^{2n+1}(3^{n-1}+1)$. I am interested in how can we strictly prove that the limit is $3$.

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Fine that you don't have the proof, yet you believe that nested radical equals $\;3\;$ : why? What have led you to believe so? –  DonAntonio Dec 31 '13 at 14:28
$3=\sqrt {1+2\cdot4}=\sqrt {1+2\cdot\sqrt {1+3\cdot5}}...$ This radical I got by similar manipulations, I know that they are informal, but I am sure in this case it is ok, and Wolfram Mathematica confirms it up to many decimal places. –  Elensil Dec 31 '13 at 14:31
And how does that relate to your radical? –  DonAntonio Dec 31 '13 at 14:32
Hrm. It sounds like if you actually showed your thoughts, you could have spared me from doing pretty much all the work I put into my answer. –  Hurkyl Dec 31 '13 at 14:34
Saying "new" is a radical move. There are more than a handful where "new" things were actually known decades, and sometimes centuries ago to people. –  Asaf Karagila Dec 31 '13 at 14:37

Let $a_0$ be your nested radical, and let $a_n$ be the radical that starts at the $n$-th term rather than at $23$.

They satisfy a recurrence relation:

$$a_0^3 - 23 = a_1$$ $$a_m^3 - 3^{2m+1}(3^{m-1} + 1) = a_{m+1}$$

Assuming $a_0 = 3$, the first few terms would be

• $a_1 = 4$
• $a_2 = 10$
• $a_3 = 28$
• $a_4 = 82$

In fact, I conjecture $a_m = 3^m + 1$ if $m > 0$.

Let $b_m = (a_m - 1)/3^m - 1$. Then, $a_m = 3^m (1+b_m) + 1$ and the recurrence relation becomes

$$(3^m (1+b_m) + 1)^3 - 3^{2m+1}(3^{m-1} + 1) = 3^{m+1} (1+b_{m+1}) + 1$$

which simplifies to

$$(3^{2m-1} (b_m^2 + 3b_m + 3) + 3^{m} (b_m + 2) + 1) b_m = b_{m+1}$$

It's clear that if

$$\lim_{x \to \infty} b_m$$

exists, then it has to equal zero. This confirms that $a_m = 3^m + 1$ for $m > 0$, and finally, that $a_0$ as conjectured.

This doesn't prove that the limit exists, though....

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If you are still interested in my thoughts : $3=\sqrt[3]{23+4}=\sqrt[3]{23+\sqrt[3]{54+10}}=\sqrt[3]{23+\sqrt[3]{54+\sqrt[3]{‌​972+28}}}$ Coefficients are exactly as i said because of the identity $(3^{n}+1)^3=3^{2n+1}(3^{n-1}+1)+3^{n+1}+1$ –  Elensil Dec 31 '13 at 14:59
$$3=\sqrt[3]{3^3}=\sqrt[3]{(3^3-4)+4}=\sqrt[3]{(3^3-4)+\sqrt[3]{4^3}}=\sqrt[3]{(3^3-4)+\sqrt[3]{(4^3-10)+10}}=\ldots,$$
where at each step the term being added and subtracted is of the form $a_{n+1}=3\cdot a_n-2$. Since $a_1$ $=4=3^1+1$, we have $a_n=3^n+1$, e.g., $a_2=10=9+1=3^2+1$. All that's left to show is that each step of the way, in order for the radical to make sense, the result of the subtraction is positive, i.e., $a_n^3-a_{n+1}>0\iff(3^n+1)^3-(3^{n+1}+1)>0\iff3^{3n}+3^{2n+1}>0$, which is indeed true for all real n.