Consider simple nested radicals
More precisely Let
$$ K > 1 , 1 \leq l $$
$$X(j,K) = X_\infty(j,K)$$
$$X_0(j,K) = a(j,K)$$
$$X_n(j,K) = \sqrt[k]{j + X_{n-1}(j,K)}$$
$$Y(j,K) = \frac{j + X_{\infty}(j,K) }{X_{\infty}(j,K) }$$
Ofcourse we have $X_{\infty}(j,K) ^k = j + X_{\infty}(j,K)$.
Then we have the limit as $n$ goes to infinity
$$ \lim_{n \to \infty} (X_{\infty}(j,K) - X_n(j,K))(Y(j,k) k)^n = C(j,K)$$
And as $K$ goes to infinity
$$C(j,\infty) = (j+1) \ln(\frac{j+1}{j})$$
——
Perhaps a proof sketch of the last statement
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 $c_{k,n} = (\alpha_k - f_k^{n-1}(1))(f_k'(\alpha_k))^{-n}$ for $n \ge 1$.
Now 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 $C_k \sim c_{k,1} \to 2\log 2$
For the more general case, we start from $x_1 = l^{1/k} = 1 + \log(l)/k + \ldots$, while $\alpha = 1 + \log(l+1)/k + \ldots$, which are again close to each other.
$\alpha - l^{1/k} \sim \log(\frac{l+1}l)/k$
$f'(\alpha) = \alpha/k(l+\alpha) \sim 1/(l+1)k$
$c_1 = (1/2)(\alpha - l^{1/k})/f'(\alpha) \sim \frac{l+1}2\log(\frac{l+1}l)$
Since $f'(\alpha^{1/k})$ and $f''(\alpha^{1/k})$ are on the order of $1/k$, we have $C = (l+1) \log(\frac{l+1}l)$
——
Now I wonder about
$$ T(j,K) = ( C(j,K) - C(j,K-1) ) ^{-2} $$
How does $T(j,K) $ behave for large $K$ ? I assume we have a good approximation of the form
$$ T(j,K) = v_1(j) + v_2(j) v_3(j)^K v_4(j)^{v_3(j)^K} $$
Where $v_i(j)$ are constants depending only on $j$.
Is this true and what are (some of) the closed forms of the v_i ??
Notice for $j = 1 $ this is related to the Paris constant.
Are there closed forms for -say- $v_2(1)$ or $ v_3(1) $ or are they as “ hopeless “ as the Paris constant itself ?
I believe in the past I have asked somewhat similar questions imho , and Taylor series were useful. In particular I mean this here
Not sure if that is relevant or helpful. I link it anyway.
