Find $\lim_{n\to\infty}\sqrt{6}^{\ n}\underbrace{\sqrt{3-\sqrt{6+\sqrt{6+\dotsb+\sqrt{6}}}}}_{n\text{ square root signs}}$

We have the following representation of pi: $$\pi=\lim_{n\to\infty}2^n \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dotsb+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}}}}_{n\text{ square root signs}}$$ which can be proven using the identity $\sin\left(\dfrac\pi{2^{n+1}}\right)=\dfrac12\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dotsb+\sqrt{2}}}}}_{n\text{ square root signs}}$. (There's similar one for $\cos$, except without the minus sign.)

This made me wonder: Is there a closed form for: $$\lim_{n\to\infty}3^n \underbrace{\sqrt{3-\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\dotsb+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}}}}}}_{n\text{ square root signs}}$$ (Note that $\sqrt{6+\sqrt{6+\sqrt{\dots}}}=3$, so this question is of the form $\infty\times0$.)

EDIT: It seems like that doesn't converge, but this does: $$\lim_{n\to\infty}\sqrt{6}^{\ n}\underbrace{\sqrt{3-\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\dotsb+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}}}}}}_{n\text{ square root signs}}\approx4.49377$$

I did not find a closed form, but I show existence of the limit, and find a very good correction for the error at a finite termination.

Convergence

Define $a_0=0$ and $$ a_{n+1}=\sqrt{6+a_n}\tag{1} $$ Then $$ \begin{align} \frac{3-a_{n+1}}{3-a_n} &=\frac{3-\sqrt{6+a_n}}{3-a_n}\\ &=\frac1{3+\sqrt{6+a_n}}\\ &\le\frac1{3+\sqrt6}\tag{2} \end{align} $$ Note that $(2)$ implies that $a_n$ increases monotonically from $0$ to $3$.

Inductively, we get $$ |\,3-a_n\,|\le\frac3{(3+\sqrt6)^n}\tag{3} $$ Therefore, $$ \begin{align} \frac{6^{n+2}}{6^{n+1}}\frac{3-a_{n+1}}{3-a_n} &=\frac6{3+\sqrt{6+a_n}}\\ &=1+\frac{3-\sqrt{6+a_n}}{3+\sqrt{6+a_n}}\\ &=1+\frac{3-a_n}{(3+\sqrt{6+a_n})^2}\tag{4} \end{align} $$ From $(3)$, we get $$ \left|\frac{3-a_n}{(3+\sqrt{6+a_n})^2}\right|\le\frac3{(3+\sqrt6)^{n+2}}\tag{5} $$ Therefore, $(4)$ and $(5)$ imply that $$ \begin{align} A &=\lim_{n\to\infty}6^{n+1}(3-a_n)\\ &=18\prod_{k=0}^\infty\frac{6^{k+2}}{6^{k+1}}\frac{3-a_{k+1}}{3-a_k}\\ &=18\prod_{k=0}^\infty\left(1+\frac{3-a_k}{(3+\sqrt{6+a_k})^2}\right)\tag{6} \end{align} $$ converges, and the limit in the question is $\sqrt{A}$.

Some Estimates

As $a_n\to3$, $3+\sqrt{6+a_n}\to6$; therefore, we have $$ \begin{align} \frac1{3+\sqrt{6+a_k}}-\frac16 &=\frac{3-\sqrt{6+a_k}}{6(3+\sqrt{6+a_k})}\\ &=\frac{3-a_k}{6(3+\sqrt{6+a_k})^2}\\ &\le\frac{3-a_k}{6(3+\sqrt6)^2}\tag{7} \end{align} $$ and using $(7)$, we get $$ \begin{align} \frac{36}{(3+\sqrt{6+a_k})^2}-1 &=36\left(\frac1{3+\sqrt{6+a_k}}-\frac16\right)\left(\frac1{3+\sqrt{6+a_k}}+\frac16\right)\\ &\le36\frac{3-a_k}{6(3+\sqrt6)^2}\frac2{3+\sqrt6}\\ &=\frac{12(3-a_k)}{(3+\sqrt6)^3}\tag{8} \end{align} $$ For $k\ge n$, $(4)$ and $(2)$ say that $$ \begin{align} \frac{6^{k+1}(3-a_k)}{6^{n+1}(3-a_n)} &=\prod_{j=n}^{k-1}\left(1+\frac{3-a_j}{(3+\sqrt{6+a_j})^2}\right)\\ &\le\exp\left(\sum_{j=n}^{k-1}\frac{3-a_j}{(3+\sqrt{6+a_j})^2}\right)\\ &\le\exp\left(\sum_{j=n}^\infty\frac{3-a_n}{(3+\sqrt6)^{j-n+2}}\right)\\ &=\exp\left(\frac{(3-a_n)3/\sqrt6}{(3+\sqrt6)^2}\right)\tag{9} \end{align} $$ Putting together $(8)$ and $(9)$, for $k\ge n$, we have $$ 1\le\frac{6^{k+1}(3-a_k)}{6^{n+1}(3-a_n)}\frac{36}{(3+\sqrt{6+a_k})^2}=1+O\left(3-a_n\right)\tag{10} $$ where the lower bound is assured by $(4)$ and $(2)$.

Correction to Finite Termination

From $(4)$, we have $$ A=6^{n+1}(3-a_n)\prod_{k=n}^\infty\left(1+\frac{3-a_k}{(3+\sqrt{6+a_k})^2}\right)\tag{11} $$ Esimating the product in $(11)$ using $(10)$, we have $$ \begin{align} &\prod_{k=n}^\infty\left(1+\frac{3-a_k}{(3+\sqrt{6+a_k})^2}\right)\\ &=\prod_{k=n}^\infty\left(1+\frac{3-a_n}{6^{k-n+2}}\frac{6^{k+1}(3-a_k)}{6^{n+1}(3-a_n)}\frac{36}{(3+\sqrt{6+a_k})^2}\right)\\ &=\prod_{k=n}^\infty\left(1+\frac{3-a_n}{6^{k-n+2}}(1+O(3-a_n))\right)\\ &=\exp\left(\sum_{k=n}^\infty\left[\frac{3-a_n}{6^{k-n+2}}+O\left(\frac{(3-a_n)^2}{6^{k-n+2}}\right)\right]\right)\\ &=\exp\left(\frac{3-a_n}{30}\right)\left(1+O(3-a_n)^2\right)\tag{12} \end{align} $$ Thus, although we can get from $(4)$ that $$ A=6^{n+1}(3-a_n)+O(3-a_n)\tag{13} $$ we get from $(12)$ that $$ A=\exp\left(\frac{3-a_n}{30}\right)6^{n+1}(3-a_n)+O(3-a_n)^2\tag{14} $$ Therefore, the correction in $(14)$ essentially doubles the number of digits known for $A$.

Example $$ \begin{align} e^{(3-a_{20})/60}&=1.000000000000000015342412876422776327920\\[4pt] \sqrt{6^{21}(3-a_{20})}&=4.493767376985934399460882507579755317836\\ e^{(3-a_{20})/60}\sqrt{6^{21}(3-a_{20})}&=4.493767376985934468406116975897359789186\\ \lim_{n\to\infty}\sqrt{6^{n+1}(3-a_n)}&=4.493767376985934468406116975897362358095 \end{align} $$ The error correction adds about $17$ correct decimal places to the estimate for $n=20$.

This isn't an answer since I have no proof, but numerically it appears that

$$ \lim_{n\to\infty} 6^{n/2} \underbrace{\sqrt{3-\sqrt{6+\sqrt{6+\dotsb+\sqrt{6}}}}}_{n\text{ square root signs}} = 1.83457284939678559111564141519063\ldots $$

Here's the Mathematica code I used to calculate this:

s[n_] := Module[{start = Sqrt[6]}, For[j = 1, j < n, j++, start = Sqrt[6 + start] ]; Return[start] ]; m = 50; 6^(m/2) Sqrt[3 - N[s[m], 100]]

This completes Slade's idea: For $t>2$, let $k=t^2-t$ and $$b_n=\underbrace{\sqrt{k+\sqrt{k+\cdots+\sqrt{k}}}}_{n}.$$ Then $b_n\uparrow t$. We are interested in the sequence $$a_n=(2t)^n(t-b_n).$$ Slade pointed out that $$a_n-a_{n-1}=(2t)^{-n-1}a_n^2.\tag{1}$$ On the other hand, $$b_n^2=k-b_{n-1}=t^2-t+b_{n-1}.$$ That gives $$t-b_n=(t-b_{n-1})\frac{1}{t+b_{n}}<\frac{t-b_{n-1}}{t}.$$ So $t-b_n<Ct^{-n}$ for some constant $C$. That means $$a_n<C2^n.\tag{2}$$ Substitute (2) to the RHS of (1) we get $$a_n-a_{n-1}<C^2(2t)^{-n-1}2^{2n}=\frac{C}{t}(t/2)^{-n}.$$ Since $t>2$, it follows that $a_n$ is bounded.

Note: Since we obtained (2) by a very crude estimation, we can improve the lower bound for $t$ a lot. I believe with a careful computation that can be pushed to $1$.

Note 2: I just realized that the question is for the actual limit, which is not obvious (if possible) from this calculation.

Here are some useful calculations. I might turn them into a solution later, but I've run out of time for the moment, and thought I'd post them as-is.

Fix $t>1$, and set $k=t^2 - t$.

Let $a_n = (2t)^n (t - \underbrace{\sqrt{k + \sqrt{k + \cdots}})}_{n}$. We have $a_0 = t$, and the following recurrence:

$$a_n - a_{n-1} = (2t)^{-n-1}a_n^2$$

So the sequence is increasing. If we can show that it is bounded, then the above equation implies that the differences shrink exponentially, so $\lim_{n\to\infty} a_n$ exists.

I'm only addressing the question of convergence here.

This is probably off by one or two factors of $6$, but I think it gets the idea across:

$$\begin{align} {1\over a_n^2}&={1\over6^n}{1\over3-\sqrt{6+\sqrt{6+\cdots}}}\\ \\ &={1\over6^n}{3+\sqrt{6+\sqrt{6+\cdots}}\over3^2-(6+\sqrt{6+\cdots})}\\ \\ &={1\over6^n}{3+\sqrt{6+\sqrt{6+\cdots}}\over3-\sqrt{6+\cdots}}\\ \\ &={1\over6^n}\left(3+\sqrt{6+\sqrt{6+\cdots}} \right)\left(3+\sqrt{6+\cdots} \right)\cdots\left(3+\sqrt{6} \right)\\ \\ &=\left(6-(3-\sqrt{6+\sqrt{6+\cdots}})\over6 \right)\left(6-(3-\sqrt{6+\cdots})\over6 \right)\cdots\left(6-(3-\sqrt{6})\over6 \right)\\ \\ &=\left(1-{3-\sqrt{6+\sqrt{6+\cdots}}\over6}\right)\left(1-{3-\sqrt{6+\cdots}\over6}\right)\cdots\left(1-{3-\sqrt{6}\over6}\right) \end{align}$$

So

$${1\over L^2}=\left(1-{3-\sqrt{6}\over6}\right)\left(1-{3-\sqrt{6+\sqrt6}\over6}\right) \left(1-{3-\sqrt{6+\sqrt{6+\sqrt6}}\over6}\right)\cdots$$

The convergence of the infinite product depends on how quickly $\sqrt{6+\sqrt{6+\cdots}}$ approaches $3$. But

$$\begin{align} 3-\sqrt{6+\sqrt{6+\cdots}} &={3-\sqrt{6+\cdots}\over3+\sqrt{6+\sqrt{6+\cdots}}}\\ \\ &=\left({1\over3+\sqrt{6+\sqrt{6+\cdots}}}\right) \left({1\over3+\sqrt{6+\cdots}}\right)\cdots \left({1\over3+\sqrt{6}}\right)\\ \\ &\lt\left({1\over3}\right)\left({1\over3}\right)\cdots\left({1\over3}\right)={1\over3^n} \end{align}$$

which is enough to show the product converges.