TITLE: How find the $\sqrt[3]{11+4\sqrt[3]{14+10\sqrt[3]{17+18\sqrt[3]{20+28\sqrt[3]{23+\cdots}}}}}$
QUESTION [14 upvotes]: find the value

$$\sqrt[3]{11+4\sqrt[3]{14+10\sqrt[3]{17+18\sqrt[3]{20+28\sqrt[3]{23+\cdots}}}}}\cdots (1)$$

It is well kown this value
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$
But for $(1)$ I can't find it.Thank you

REPLY [8 votes]: We have to look at the sequences recursively defined by $x_{n+1} = f_n(x_n)$ where $f_n(x) = \frac{x^3-3n-8}{n(n+3)}$ and $x_1 \in \Bbb R$.
As did MvG, we can check that one such sequence is given by $u_n = n+2$, and we have to prove how special this sequence is :
If $x_1 > u_1$, then $x_n > u_n$ (because the $f_n$ are increasing). Since $f_n'$ is also increasing and $f'_n(u_n) = 3u_n^2/n(n+3) \to 3$, there exists an $n_0$ such that $|x_{n+1}-u_{n+1}| > 2|x_n-u_n|$ for $n \ge n_0$. Hence $(x_n)$ explodes at least exponentially.
If $x_1 < u_1$, then $x_n < u_n$ and eventually, $x_n < n$, because around $u_n$ and for $n$ large enough, the difference grows exponentially (there is a small region around $n$ and $u_n$ where $f'n(x_n)$ stays arbitrarily close to $3$ as long as $n$ is large enough).
Now, since for positive $x_n$, $x_{n+1}/(n+1) < (x_n/n)^3$ and since you quickly end up near $0$ when you cube iteratively, the sequence $x_n/n$ is decreasing and must get arbitrarily close to $0$. It does so pretty fast ($x_n \le a^{3^n}n$ for some $a<1$), and eventually, $x_n$ will get below $\sqrt[3]{3n+8}$ and the sequence will turn negative.
This proves that the truncated sequence $\sqrt[3]{11}, \sqrt[3]{11+4 \sqrt[3]{14}}, \ldots$converge to $3$ : if you write only $n$ roots, you are looking at the $x_1^{(n)}$ making $x_{n+1}=0$. Since $u_n > 0$, we have $x_1^{(n)} < u_1$. That sequence $x_1^{(n)}$ is increasing and has some limit $l \le u_1$. Since the sequence obtained starting at $l$ is positive, we must have $l=u_1$.

It seems that for $0 \le u_1 < 3$, the sequence converge to $0$, while for large negative $u_1$, the sequence obviously diverges to $ - \infty$ very quickly. So somewhere there should be another critical (negative) value for $x_1$ whose generated sequence sits in-between those converging to $0$ and those that are exploding.