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 $3$.
On
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.
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
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....