Conjecture: $\lim\limits_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots )^{-1/x}\stackrel?=e$

Question

Well, it's a conjecture so let me propose it:

$$\lim_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots)^{-1/x}\stackrel?=e$$

Where I use desmos notation and $x!! := ((x!)!,x!!!=(((x!)!)!)$

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It seems so hard that I haven't any clue to show it. I already know that

$$x!^{\frac{-1}{x}}=e^{\gamma}$$

Perhaps we can use the Weierstrass factorization theorem and compute it. So, how to (dis)prove it?

Further investigation :

If we supposed that the following functions are convex on $(0,1)$:

$$a_1(x)=x!,a_2=x!!,\cdots,a_n=x!!\cdots !$$

Then we can rewrite the conjectured limit as :

$$L=\lim_{x\to 0}\left((a_1'(0)x+1)(a_2'(0)x+1)\cdots(a'_n(0)x+1)\cdots\right)^{\frac{-1}{x}}=^?e$$

Where :

$$a_1'(0)=\gamma,a_2'(0)=(\gamma-1)\gamma,a_3'(0)=-(\gamma-1)^2\gamma,a_4'(0)=(\gamma-1)^3\gamma,\cdots\tag{I}$$

Update :

It seems we have :

$$\lim_{x\to 0^+}\left(x!x!!x!!!x!!!!x!!!!!...\right)^{\frac{2^{x}-1}{x^{2}}}=1/2$$

Update $2$ :

Using $I$ and the fact that (see Robjohn's answer) :

$$\sum_{n=1}^\infty\gamma(1-\gamma)^{n-1}=1$$

And :

Let $x_i\in[1-1/n,1]$ where $n\geq M$ two natural numbers large enought then we have :

$$\sum_{i=1}^{n}x_i-(n-1)\leq \prod_{i=1}^{n}x_i\leq \sum_{i=1}^{n}x_i-(n-1)+\frac{1}{2n}$$

We have after simplification :

$$\left(-x+1\right)^{\left(-\frac{1}{x}\right)}<L<\left(-x+1+\frac{1}{2n}\right)^{\left(-\frac{1}{x}\right)}$$

now let $n\to \infty$ and $x\to 0$ we get the result .

Ps:It's a try and I think it should be clearing a little (the LHS seems dubious) and the credit come back to @Robjohn.

Answer 1

We will work with $x\ge0$ because the bounds work out cleaner, but to extend to $x\le0$, we simply need to increase $a_1$ and $\beta$ a bit.

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Developing $\pmb{g_n}$

Define $g_0(x)=x$ and $$ g_n(x)=\Gamma(1+g_{n-1}(x))\tag1 $$ First, for $x\approx0$, we have $\Gamma'(1)=-\gamma$, giving $$ g_1(x)=1-\gamma x+[0,a_1]_\#x^2\tag2 $$ where $a_1=\frac{\Gamma''(1)}2$, using $\Gamma'''(1)\lt0$. We use the notation $[a,b]_\#$ to mean a real number in $[a,b]$.

Next, the value of $g_1(x)$ is near $1$ so the argument of the outer $\Gamma$ in $g_2(x)=\Gamma(1+\Gamma(1+x))$ is $\approx2$, and $\Gamma'(2)=1-\gamma$. Therefore, $$ g_2(x)=1-\gamma(1-\gamma)x+[0,a_2]_\#x^2\tag3 $$ where $a_2=\underbrace{(1-\gamma)a_1}_{\substack{\text{contribution}\\\text{from the}\\\text{$2^\text{nd}$ order}\\\text{term in $(2)$}}}+\underbrace{\quad\beta\gamma^2\quad}_{\substack{\text{contribution}\\\text{from the}\\\text{$1^\text{st}$ order}\\\text{term in $(2)$}}}$, and $\beta=\frac{\Gamma''(2)}2$, using $\Gamma'''(2)\gt0$.

Repeating this process gives $$ g_n(x)=1-\gamma(1-\gamma)^{n-1}x+[0,a_n]_\#x^2\tag4 $$ where $$ a_n=(1-\gamma)a_{n-1}+\beta\gamma^2(1-\gamma)^{2n-4}\tag5 $$

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Solving the recurrence for $\pmb{a_n}$

Letting $a_n=b_n(1-\gamma)^n$, we have $b_1=\frac{a_1}{1-\gamma}$ and $(5)$ becomes $$ \begin{align} b_n &=b_{n-1}+\beta\gamma^2(1-\gamma)^{n-4}\tag{6a}\\[9pt] &=\frac{a_1}{1-\gamma}+\frac{\beta\gamma^2}{(1-\gamma)^2}\frac{1-(1-\gamma)^{n-1}}{1-(1-\gamma)}\tag{6b}\\ &=\frac{a_1(1-\gamma)+\beta\gamma\left(1-(1-\gamma)^{n-1}\right)}{(1-\gamma)^2}\tag{6c} \end{align} $$ Therefore, $$ \begin{align} a_n &=\left(a_1(1-\gamma)+\beta\gamma\left(1-(1-\gamma)^{n-1}\right)\right)(1-\gamma)^{n-2}\tag{7a}\\[3pt] &\le(a_1(1-\gamma)+\beta\gamma)(1-\gamma)^{n-2}\tag{7b}\\[3pt] &=\eta\,(1-\gamma)^{n-2}\tag{7c} \end{align} $$ where $\eta=a_1(1-\gamma)+\beta\gamma=\frac{\pi^2}{12}-\frac{\gamma^2}2$.

Plugging $(7)$ into $(4)$ gives $$ g_n(x)=1-\gamma(1-\gamma)^{n-1}x+\left[0,\eta\,(1-\gamma)^{n-2}\right]_\#x^2\tag8 $$

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Evaluating the Product

Since $\frac{\gamma(1-\gamma)}\eta\ge\frac13$, if $x\in\left[0,\frac13\right]$, then $g_n(x)\le1$; if $x\in\left[-\frac13,0\right]$, then $g_n(x)\ge1$. In either case, we can apply the Theorem from this answer. To apply the Theorem, note that $$ \sum_{n=1}^\infty\gamma(1-\gamma)^{n-1}=1\tag9 $$ and $$ \sum_{n=1}^\infty\eta(1-\gamma)^{n-2}=\frac\eta{\gamma(1-\gamma)}\le3\tag{10} $$ Applying the Theorem, using $(9)$ and $(10)$, gives $$ 1-x\le1-x+[0,3]_\#x^2\le\prod_{n=1}^\infty g_n(x)\le\frac1{1+x-[0,3]_\#x^2}\le1-x+4x^2\tag{11} $$ That is, $$ \prod_{n=1}^\infty g_n(x)=1-x+[0,4]_\#x^2\tag{12} $$

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The Root of the Matter

Using the result of this answer, we can compute $$ \begin{align} \lim_{x\to0}\left(\prod_{n=1}^\infty g_n(x)\right)^{-1/x} &=\lim_{x\to0}\left(1-x+[0,4]_\#x^2\right)^{-1/x}\tag{13a}\\[3pt] &=e\tag{13b} \end{align} $$

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$\pmb{\eta}$ Seems to be Correct

Using Mathematica and the functions

g[n_,x_]:=N[Nest[Gamma[1+#]&,x,n],100]

and

f[n_,x_]:=(g[n,x]-(1-EulerGamma(1-EulerGamma)^(n-1)x))/
          ((1-EulerGamma)^(n-2)x^2)

I evaluated f[100,1/1000000000000] and got $$ \left.\frac{g_n(x)-\left(1-\gamma(1-\gamma)^{n-1}x\right)}{(1-\gamma)^{n-2}x^2}\right|_{\substack{n=100\ \ \ \\x=10^{-12}}}=\color{#C00}{0.655878071518}898822 $$ whereas $\eta=\color{#C00}{0.655878071520}253881$, a difference of ${}\approx1.355\times10^{-12}$.

Answer 2

Let $f_0=x$ and $f_n=\big[f_{n-1}\big]!$ and

$$P_k=\Big[\prod_{n=1}^k f_n\Big]^{-\frac 1 x}$$ Take logarithm and use series expansions to get $$\left( \begin{array}{ccc} k & \log(P_k) & \text{numerical value} \\ 1 & \gamma & 0.577216 \\ 2 & 2 \gamma -\gamma ^2 & 0.821253 \\ 3 & 3 \gamma -3 \gamma ^2+\gamma ^3 & 0.924429 \\ 4 & 4 \gamma -6 \gamma ^2+4 \gamma ^3-\gamma ^4 & 0.96805 \\ 5 & 5 \gamma -10 \gamma ^2+10 \gamma ^3-5 \gamma ^4+\gamma ^5 & 0.986492 \\ 6 & 6 \gamma -15 \gamma ^2+20 \gamma ^3-15 \gamma ^4+6 \gamma ^5-\gamma ^6 & 0.994289 \\ 7 & 7 \gamma -21 \gamma ^2+35 \gamma ^3-35 \gamma ^4+21 \gamma ^5-7 \gamma ^6+\gamma ^7 & 0.997585 \\ 8 & 8 \gamma -28 \gamma ^2+56 \gamma ^3-70 \gamma ^4+56 \gamma ^5-28 \gamma ^6+8 \gamma ^7-\gamma ^8 & 0.998979 \\ 9 & 9 \gamma -36 \gamma ^2+84 \gamma ^3-126 \gamma ^4+126 \gamma ^5-84 \gamma ^6+36 \gamma ^7-9 \gamma ^8+\gamma ^9 & 0.999568 \end{array} \right)$$ where simple and interesting patterns appear $(\large !!)$

Continuing withe the numerical values $$\left( \begin{array}{cc} k & \text{numerical value} \\ 10 & 0.99981753109089894176 \\ 11 & 0.99992285500358956645 \\ 12 & 0.99996738430398644122 \\ 13 & 0.99998621059464713383 \\ 14 & 0.99999417005542648523 \\ 15 & 0.99999753519075982564 \\ 16 & 0.99999895791726424833 \\ 17 & 0.99999955942374344764 \\ 18 & 0.99999981373126031334 \\ 19 & 0.99999992124849474194 \\ 20 & 0.99999996670509721147 \\ 21 & 0.99999998592343666238 \\ 22 & 0.99999999404864952883 \\ 23 & 0.99999999748386224811 \\ 24 & 0.99999999893621637355 \\ 25 & 0.99999999955024894680 \\ 26 & 0.99999999980985230007 \\ 27 & 0.99999999991960853110 \\ 28 & 0.99999999996601174633 \\ 29 & 0.99999999998563029871 \\ 30 & 0.99999999999392471546 \end{array} \right)$$

Apparently, you have found a new definition of $\color{red}{\large 1}$.

Edit

$\log(P_{250})=1-3.39 \times 10^{-94}$

Answer 3

At first Mathematica appears to support your conjecture numerically, e.g. using $x!=\Gamma(x+1)$ then $$\underset{x\to 0}{\text{lim}}(\Gamma (x+1) \Gamma (\Gamma (x+1)+1) \Gamma (\Gamma (\Gamma (x+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1)+1)+1) \Gamma (\Gamma (\Gamma (\Gamma (\Gamma (\Gamma (\Gamma (x+1)+1)+1)+1)+1)+1)+1))^{-1/x}$$ You can try it yourself if you have Mathematica (thanks to @KStarGamer for the improved code):

Limit[Product[Nest[Gamma[# + 1] &, x, n], {n, 1, 7}]^(-1/x), x -> 0]

In the above example we obtain $$\exp \left(\gamma \left(7-21 \gamma +35 \gamma ^2-35 \gamma ^3+21 \gamma ^4-7 \gamma ^5+\gamma ^6\right)\right),$$ and the exponent does seem to tend to $1$.

The coefficients in the exponent polynomial in $\gamma$ appear to be those of A007318 (binomial coefficients !) So the general case could be binomial in some way.

It looks like the general case is

$$e^{1-(1-\gamma )^n}\to e^1$$ as $n\to\infty$.

For a proof you may be able to use the fact that $$\Gamma(x+1)=1-\gamma x + O(x^2).$$

Answer 4

If we consider the function $$f(x) = \sum_{k=1}^\infty \log(x(!^k))$$ then the question is equivalent to prove that $f'(0)=-1$.

We have (assuming that we can differentiate term by term) that$$f'(x) = \sum_{k=1}^\infty\frac{d}{dx} \log(x(!^k))= \sum_{k=1}^\infty\frac{\frac{d}{dx} x(!^k)}{x(!^k)}$$

Since $0(!^k)=1$ we have

$$f'(0)=\sum_{k=1}^\infty \left.\frac{d}{dx}x(!^k)\right\vert_{x=0}$$

Using $\frac{d}{dx}x! = x!\psi(1+x)$ and a repeated application of the chain rule yields $$\frac{d}{dx}(x(!)^k) = x! x(!^2)x(!^3)\cdots x(!^k) \psi(1+x)\psi(1+x(!^1))\psi(1+x(!^2))\cdots\psi(1+x(!^{k-1}))$$

where $\psi$ is the digamma function.

Using again that $0(!^k) = 1$, that $\psi(1) = -\gamma$ and $\psi(2) = 1-\gamma$ we see that $\left.\frac{d}{dx}x(!^k)\right\vert_{x=0} = -\gamma\,(1-\gamma)^{k-1}$ and then we have $$f'(0)=\sum_{k=1}^\infty-\gamma\, (1-\gamma)^{k-1} =-\gamma\,\frac{1}{1-(1-\gamma)} = -1$$