# A limit related to super-root (tetration inverse).

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.

$$L \ln2 = r = 0.0475…$$

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