TITLE: Infinite radical. How to show it converges to three?
QUESTION [7 upvotes]: 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 $3$.

REPLY [2 votes]: Not quite sure what's there to ‘prove’, really. What you wrote basically boils down to:
$$3=\sqrt[3]{3^3}=\sqrt[3]{(3^3-4)+4}=\sqrt[3]{(3^3-4)+\sqrt[3]{4^3}}=\sqrt[3]{(3^3-4)+\sqrt[3]{(4^3-10)+10}}=\ldots,$$
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.