On the Paris constant and $\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\dots}}}}$? In 1987, R. Paris proved that the nested radical expression for $\phi$,
$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
approaches $\phi$ at a constant rate. For example, defining $\phi_n$ as using $n = 5, 6, 7$ "ones" respectively, then,
$$(1/2)(\phi-\phi_5)(2\phi)^5 = 1.0977\dots$$
$$(1/2)(\phi-\phi_6)(2\phi)^6 = 1.0983\dots$$
$$(1/2)(\phi-\phi_7)(2\phi)^7 = 1.0985\dots$$
which is approaching the Paris constant $R = 1.09864196\dots$. It seems it can be generalized. Define,
$$x_n=\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1_n+\dots}}}}\tag{1}$$
for integer $k>1$ and the equations,
$$x^k = x+1\tag{2}$$
$$y = \frac{1}{x}+1\tag{3}$$ 
where $x$ is the root of $(2)$ such that $x = x_n$ as $n \to \infty$ in $(1)$.  Then one can conjecture that,
$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_k\tag{4}$$
for some constant $C_k$. The Paris constant is simply the case $C_2$. 
I tested it for increasing large $k$. The sequence of $C_k$ seem to be themselves approaching a constant. The rate is very slow, so for much higher $k = 10^{14},10^{15},10^{16}$,
$$C_{10^{14}} =  0.6931471805599500\dots$$
$$C_{10^{15}} =  0.6931471805599457\dots$$
$$C_{10^{16}} =  0.6931471805599454\dots$$
Compare to,
$$\ln 2 =  0.6931471805599453\dots$$
Question:


*

*Does $C_k \to \ln 2$, as $k \to \infty$? 



$\color{blue}{Edit,\; Nov.\;25}$
More generally, define,
$$x_n=\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a+\sqrt[k]{a_n+\dots}}}}\tag{5}$$
for integers $a\ge 1,\;k>1$ and,
$$x^k = x+a\tag{6}$$
$$y = \frac{a}{x}+1\tag{7}$$ 
Then it seems,
$$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}\tag{8}$$
The Paris constant is the case $C_{1,2}$. Is it true that as $k \to \infty$, then,
$$\lim_{k\to \infty} C_{1,k} =  \ln 2$$
$$\lim_{k\to \infty} C_{2,k} = \tfrac{3}{2} \ln \tfrac{3}{2}$$
$$\lim_{k\to \infty} C_{3,k} = \tfrac{4}{2} \ln \tfrac{4}{3}$$
$$\lim_{k\to \infty} C_{4,k} = \tfrac{5}{2} \ln \tfrac{5}{4}$$
and so on?
P.S. The only known closed-form in terms of transcendental constants is $C_{2,2} = \pi^2/8$.
 A: Antonio Vargas's observation means that $1$ starts closer and closer to the fixpoint, so that maybe there is less and less difference between $C_k$ and the first term in the sequence defining it ; and maybe that first term converges to $\log 2$.
Let $f_k(x) = \sqrt[k]{1+x}$ for $x \ge 0$ and $k > 1$.
Let $\alpha_k$ the unique positive fixpoint  of $f_k$ (it is the positive root of $\alpha_k^k = \alpha_k+1$).
Define $c_{k,n} = (1/2)(\alpha_k - f_k^{n-1}(1))(f_k'(\alpha_k))^{-n}$ for $n \ge 1$.
Now your constant $C_k$ is defined by $C_k = \lim_{n \to \infty} c_{k,n}$, and we want to give an estimation of $C_k/c_{k,1}$. 
We have $c_{k,{n+1}}/c_{k,n} = f_k'(\alpha_k)^{-1}(f_k(\alpha_k) - f_k(f_k^{n-1}(1)))/(\alpha_k - f_k^{n-1}(1)) = f_k'(\alpha_k)^{-1}f_k'(z_{k,n})$ for some $f_k^{n-1}(1) \le z_{k,n} \le \alpha_k$.
Since $f_k'$ is decreasing, we obtain 
 $1 \le c_{k,n+1}/c_{k,n} \le f_k'(f_k^{n-1}(1))f_k'(\alpha_k)^{-1}$
Some crude estimates gives us $\alpha_k \ge f_k^n(1) \ge \alpha_k - (\alpha_k - 1)f_k'(1)^n$,
and then ($f''_k$ is increasing), $f'_k(\alpha_k) \le f_k'(f_k^n(1)) \le f_k'(\alpha_k) -  (\alpha_k-1)f'_k(1)^nf_k''(1)$,
and finally $1 \le c_{k,n+1}/c_{k,n} \le 1 + (\alpha_k-1)f'_k(1)^{n-1}(-f''_k(1))f_k'(\alpha_k)^{-1} $.
Using $1+x \le \exp(x)$ and taking the product,
we obtain $1 \le C_k/c_{k,1} \le \exp((\alpha_k-1)(-f''_k(1))f'_k(\alpha_k)^{-1}/(1-f'_k(1))) $

Since $\alpha_k = 1 + \log 2/k + O(k^{-2})$, we have  
$\alpha_k-1 \sim \log2 / k$
$f'_k(\alpha_k) = \alpha_k(1 + \alpha_k)^{-1}/k \sim 1/2k$
$c_{k,1} = (1/2)(\alpha_k-1)f'_k(\alpha_k)^{-1} \sim (1/2)(\log 2/k)(2k) = \log 2$
$f_k'(1) = 2^{1/k-1} \frac 1k \sim 1/2k$
$f''_k(1) = 2^{1/k-2} \frac 1k (\frac 1k -1) \sim -1/4k$
$(\alpha_k-1)(-f''_k(1))f'_k(\alpha_k)^{-1}/(1-f'_k(1)) \sim \log 2/2k \to 0$
This shows that $C_k \sim c_{k,1} \to \log 2$

For the more general case, we start from $x_1 = a^{1/k} = 1 + \log(a)/k + \ldots$, while $\alpha = 1 + \log(a+1)/k + \ldots$, which are again close to each other.
$\alpha - a^{1/k} \sim \log(\frac{a+1}a)/k$
$f'(\alpha) = \alpha/k(a+\alpha) \sim 1/(a+1)k$
$c_1 = (1/2)(\alpha - a^{1/k})/f'(\alpha) \sim \frac{a+1}2\log(\frac{a+1}a)$
Since $f'(\alpha^{1/k})$ and $f''(\alpha^{1/k})$ are on the order of $1/k$, we have $C = \frac{a+1}2\log(\frac{a+1}a)$
A: Hint : $\displaystyle x(n)=\underset{k=0}{\overset\infty{\Large\Xi}}\left(a,b\,;\tfrac1n\right)\iff x^n=a+bx\iff n(x)=\frac{\ln(a+bx)}{\ln x}\iff n(1)=\infty$ , $n(\infty)=1$ . Now show, using l'Hopital, that $\displaystyle\lim_{x\to1}\Big[n(x)\cdot(x-1)\Big]=\ln2$.
A: You might notice that if instead of:
$\lim_{n\to\infty}(1/2)(x-x_n)(ky)^n = C_{a,k}^+\tag{8}$
You don't divide by 2:
$\lim_{n\to\infty}(x-x_n)(ky)^n = C_{a,k}^+\tag { }$
The known closed form constant is then $C_{2,2}^+=\frac{pi^2}{4}$, while the other version (below) with $a=x^k+x$ has a closed form constant of $C_{2,2}^{-}=\frac{pi}{2\sqrt{3}}$.
other version:
$x_n=\sqrt[k]{a-\sqrt[k]{a-\sqrt[k]{a-\sqrt[k]{a_n-\dots}}}}\tag { }$
C+ approaches 1 from above:
$\lim_{k\to \infty} C_{1,k}^+ = 2 \ln 2\tag{}$
$\lim_{k\to \infty} C_{2,k}^+ = 3 \ln \tfrac{3}{2}\tag{}$
$\lim_{k\to \infty} C_{3,k}^+ = 4 \ln \tfrac{4}{3}\tag{}$
C- approaches 1 from below: 
$\lim_{k\to \infty} C_{2,k}^- = \ln 2 \tag{}$
$\lim_{k\to \infty} C_{3,k}^- = 2 \ln \frac{3}{2}\tag{}$
$\lim_{k\to \infty} C_{4,k}^- = 3 \ln \frac{4}{3} \tag{}$
You'll notice that the second form (with $a=x^k+x$) has a C that approaches 1 from below, and the first one (you posted about) approaches 1 from above. 
$\lim_{k\to \infty} C_{a,k}^+ = (a+1) \ln \frac{a+1}{a} \tag{}$
$\lim_{k\to \infty} C_{a,k}^- = (a-1) \ln \frac{a}{a-1} \tag{}$
Coincidentally :D $\lim_{a\to\infty} (\frac{a}{a-1})^a \to e^+$ from above, and $\lim_{a\to\infty} (\frac{a}{a-1})^{a-1} \to e^-$ from below.
