On the Paris constant and $\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\sqrt[k]{1+\dots}}}}$?

Question

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:

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

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$\color{blue}{Edit,\; Nov.\;25}$

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 $C_{2,2} = \pi^2/8$.

Answer 1

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))) $

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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 $C_k \sim c_{k,1} \to \log 2$

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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 $C = \frac{a+1}2\log(\frac{a+1}a)$

Answer 2

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 $\displaystyle\lim_{x\to1}\Big[n(x)\cdot(x-1)\Big]=\ln2$.

Answer 3

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.