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Recall that tetration ${^n}x$ for $n\in\mathbb N$ is defined recursively: ${^1}x=x,\,{^{n+1}}x=x^{({^n}x)}$.

Its inverse function with respect to $x$ is called super-root and denoted $\sqrt[n]y_s$ (the index $_s$ is not a variable, but is part of the notation — it stands for “super”). For $y>1, \sqrt[n]y_s=x$, where $x$ is the unique solution of ${^n}x=y$ satisfying $x>1$. It is known that $\lim\limits_{n\to\infty}\sqrt[n]2_s=\sqrt{2}$. We are interested in the convergence speed. It appears that the following limit exists and is positive:

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

Numerically,

$$\mathcal L\approx0.06857565981132910397655331141550655423…\tag2$$

Can we prove that the limit $(1)$ exists and is positive? Can we prove that the digits given in $(2)$ are correct? Can we find a closed form for $\mathcal L$ or at least a series or integral representation for it?

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*I have some observations, from where someone more experienced might be able to derive the proof – maybe this is helpful)*

Let the iterated functional root (“superroot of order”) $B(z,n)$ (which finds the ” ** B** “ase of the powertower) be defined as

$$ \;^n b = z \qquad \to \qquad B(z,n) = b $$

For the following let us always denote $u$ for the $\log(z)$ for notational convenience.

Let us then define the generalization of the Lambert-W-function to higher iterates as a simple conjugate of the base-finding $B()$-function:

$$ W(u,n) = W(\log(z),n) = \log( B(\exp(u),n)) \tag 1$$

where we can find a formal power series for $W(u,n)$ from the Lagrange series reversion of $u \cdot \exp(u)$, such that

$$ \begin{array}{} \mathcal {\text{ Taylor:} } & W^{-1}(u,1) = u \\

\mathcal {\text{ Taylor:} }& W^{-1}(u,2) = u \cdot \exp(u) \\

\mathcal {\text{ Taylor:} }& W^{-1}(u,3) = u \cdot \exp(u \cdot \exp(u) ) \\

\end{array} \tag 2$$

Then your limit for a general $z$ and $u = \log(z)$ reads

$$ \lim_{n \to \infty} {B(z,n)-z^{1/z} \over \log(z)^n} \qquad \text{or}

\qquad \lim_{n \to \infty} {\exp(W(u,n))-\exp(u \cdot \exp(-u)) \over u^n} \tag 3 $$

Actually we use $z=2$ (and thus $u=\log(2)$) and this reads a bit more friendly:

$$ \lim_{n \to \infty} {\exp(W(u,n))- \sqrt 2 \over u^n} \tag 4$$

Now, if we look at the formal power series for $W(u,n)$ we find a striking pattern, in that with increasing *n* the coefficients of the equivalently longer leading part become constant, which can be written like the following

$$ \begin{array} {}

W(u,n) &=& u \Large \left(1-u+{ u^2 \over 2!}-{ u^3 \over 3!} + … \right. \\

&& \qquad + u^{n-1} \Large \left( {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} + … \right) \\

&& \left. \quad \Large \right)\end{array} \tag 5$$

Let us denote the part in the inner parenthese as $R(u,n)$

$$ R(u,n) = {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} – … \tag 6

$$

which makes then

$$ \begin{array} {rrl}

W(u,n) &=& u \cdot \exp(-u) + u^n \cdot R(u,n) )\\

B(z,n) &=& \exp(u \cdot \exp(-u) + u^n \cdot R(u,n)) \\

&=& \exp(u \cdot \exp(-u)) \cdot \exp( u^n \cdot R(u,n)) \\

&=& \sqrt 2\cdot \exp( u^n \cdot R(u,n)) \\

\end{array} \tag 7$$

Something on convergence: because the first line in fomula *(5)* contains an exponential series in $u$ and such a series is entire, the consideration of range of convergence of the whole construct focuses on $R(u,n)$ – other than with $z=2$ and $u=\log(2)$ I did not check it, but for this settings it seems to converge to something $\lim_{n \to \infty} R(u,n) =r_\infty \approx 0.0484903140769$ just by numerically searching (binary search) $b_n=B(2,n)$ and computing backwards. Interestingly the coefficients of the power series of $R(u,n)$ have the same scheme to become constant for the leading part which extends when *n* increases and has thus also a limiting power series. Pari/GP gives me

$$ R(u,12)= u – 5 \cdot u^2/2! + 13 \cdot u^3/3! – 19 \cdot u^4/4! + 1 \cdot u^5/5! + 231 \cdot u^6/6! – … $$

and this is the same for all values *n* greater than 12, so we might assume this constant also for the case of $n \to \infty$.

I did not find anything related in OEIS yet, perhaps I can later find some pattern…

We see in the last formula in (7) the squareroot occuring, so we get the initial formula as

$$ \lim_{n \to \infty}\sqrt2 \cdot { \exp(u^n R(u,n)) -1 \over u^n} \tag 8 $$

where the exponentialseries (divided by $u^n$) has no constant term and the linear term is $ R(u,n)$ and the following terms are powers of it, multiplied by n’th powers of $u=\log(2) \approx 0.693 \lt 1$.

These are so far only some (organized) observations, a real proof of the existence of a finite limit must be provided by analysis of the power series of $R(u,n)$ (but besides the heuristics I don’t have the proof yet)

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

Notice the resemblance with the Koenigs function

https://en.m.wikipedia.org/wiki/Koenigs_function

In fact it is a Koenigs function with the variable fixed to the value $2$.

Since $1 < 2 < \exp(1/e)$ and the derivative is not $0$ or $1$ , the Koenigs function converges to the correct value ; your limit.

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

This limit is only possible if

$$\lim\limits_{n\to\infty}\frac{\sqrt[n+1]2_s – \sqrt 2}{\sqrt[n]2_s – \sqrt 2}= \ln2$$

To show this , use l’hopital

We get with $f(x) = x^{f(x)}$ :

$ \frac{D x^{f(x)}} {f ‘ (x)} = \frac{ \sqrt 2 ^2 2\ln(\sqrt2) }{2} = \ln2$.

Qed

This is part of the answer that justifies the RHS of the limit

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

With thanks to Tommy1729 for hints.

I got a message from Tommy1729.

He considered nonzero $T$ such that :

$$T = \lim \frac{A^n}{(f(n,2) – \sqrt 2 – L \ln 2 ^n)}$$.

Where $f(n,2)$ is the nth superroot of $2$ , $L$ is the constant from the Op and $A$ is a constant.

If the limit $T$ does not exist at least the best fitting $A$ is considered.

In other words :

$$A^n $$

~

$$\frac{1}{(f(n,2) – \sqrt 2 – L \ln 2 ^n)}$$

Generalizing the Op is easy.

And the problem was not new to him.

These generalizations seem even harder and he did not provide formal proofs.

However what amazed me was this :

$$L \ln2 = r = 0.0475…$$

Where $ r $ is conjectured the radius of Gottfried’s answer !

Sorry too add questions instead of answers. But this is too much for a comment.

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