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

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

Solution 1:

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.

Solution 2:

Numerical experiments indicate that the value should be $3$ as well. So let's try to prove that.

\begin{align*} \sqrt[3]{11 + 4\sqrt[3]{14 + 10\sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}}}} &= 3 \\ 11 + 4\sqrt[3]{14 + 10\sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}}} &= 3^3 = 27 \\ 4\sqrt[3]{14 + 10\sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}}} = 3^3 &= 27-11 = 16 \\ \sqrt[3]{14 + 10\sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}}} &= \frac{16}{4} = 4 \\ 14 + 10\sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}} &= 4^3 = 64 \\ 10\sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}} &= 64 - 14 = 50 \\ \sqrt[3]{17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}}} &= \frac{50}{10} = 5 \\ 17 + 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}} &= 5^3 = 125 \\ 18\sqrt[3]{20 + 28\sqrt[3]{23 + \dots}} &= 125 - 17 = 108 \\ \sqrt[3]{20 + 28\sqrt[3]{23 + \dots}} &= \frac{108}{18} = 6 \end{align*}

I believe you can see a pattern emerging: the nested cubic roots form an arithmetic sequence of natural numbers. You can use this to obtain a more formal proof, beyond this simple “proof by example”. But if all you need is the value, don't bother.

Note that the above sequence of equations would be technically possible for any value you assume in the first line. The key point here is the fact that the values of the cubic roots grow only slowly. This means that roots nested more deeply in the whole expression have only little impact on the outermost value.