Conjecture about the minimum of the Gamma function

Question

Problem/Conjecture:

Let the function :

$$f(x)=\frac{((x+x_{\min})!-(x_{\min})!)^{\frac{1}{x}}}{x^{\frac{1}{x^2}}}$$

Where $x_\min$ denotes the minimum abscissa of the Gamma function near by $0.4616$

Then do we have :

$$f'(1-x_{\min})<^?0$$

An computation show it could be true.

And what is :

$$\lim_{x\to \infty}(f(x)-f(x-1))=?$$

Is it true? If yes is it trivial or well-know ?

Answer 1

Using for an approximation of $x_{\text{min}}$ the rational values $$\left\{\frac{343}{743},\frac{379}{821},\frac{2691 5}{58304},\frac{47381}{102638},\frac{51171}{110848},\frac{102721}{222517},\frac{256992}{556703},\frac{1439231}{3117701}\right\}$$ what is obtained $$\left( \begin{array}{cc} n & f'(1-x_{\text{min}}) \\ 1 & -0.00011212968 \\ 2 & -0.00001826854 \\ 3 & -0.00001806921 \\ 4 & -0.00001815531 \\ 5 & -0.00001816370 \\ 6 & -0.00001816408 \\ 7 & -0.00001816416 \\ 8 & -0.00001816417 \\ \end{array} \right)$$

For the second question, using $x=10^{12}$ to avoid overflows, it is quite impressive $$\left( \begin{array}{cc} n & f(10^{12})-f(10^{12}-1) \\ 1 & 0.367879441171796089914000535729 \\ 2 & 0.367879441171796086295532027426 \\ 3 & 0.367879441171796086287846671071 \\ 4 & 0.367879441171796086291166324500 \\ 5 & 0.367879441171796086291489671976 \\ 6 & 0.367879441171796086291504586674 \\ 7 & 0.367879441171796086291507556416 \\ 8 & 0.367879441171796086291507768373 \\ \end{array} \right)$$

which is $$\huge \frac 1 e$$

Edit

For the behavior of the difference, assuming that we can neglect $s_{\text{min}}$ and that $$f(x)-f(x-1) \sim f'(x)$$ using $$f(x)= \frac{\Gamma (x+1)^{\frac{1}{x}}}{x^{\frac{1}{x^2}} }$$ $$f'(x)= \frac{\Gamma (x+1)^{\frac{1}{x}}}{x^{3+\frac{1}{x^2}} } \left(x^2 \psi ^{(0)}(x+1)+2 \log (x)-x \log (\Gamma (x+1))-1\right)$$ $$\log(f'(x))=-1+\frac 1 {2x}+O\left(\left(\frac{\log (x)}{x}\right)^2\right)$$

Answer 2

This is an answer to the second question. We have $$ ((x + x_{\min } )! - (x_{\min } )!)^{1/x} = (x + x_{\min } )!^{1/x} \left( {1 - \frac{{(x_{\min } )!}}{{(x + x_{\min } )!}}} \right)^{1/x} \\= (x + x_{\min } )!^{1/x} + \mathcal{O}\!\left( {\frac{1}{x}} \right), $$ and $$ x^{ - 1/x^2 } = 1 + \mathcal{O}\!\left( {\frac{{\log x}}{{x^2 }}} \right), $$ and $$ (x + x_{\min } )!^{1/x} \sim \frac{x}{\mathrm{e}}, $$ as $x\to +\infty$. Thus $$\tag{1} f(x) = (x + x_{\min } )!^{1/x} + \mathcal{O}\!\left( {\frac{{\log x}}{x}} \right) $$ Now, for any fixed $a$, $$ \Gamma (z + a) = z^{z + a - 1/2} \mathrm{e}^{ - z} \sqrt {2\pi } \left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right) $$ as $z\to +\infty$. Consequently, \begin{align*} (\Gamma (z + a))^{1/z} & = z^{1 + a/z - 1/(2z)} {\rm e}^{ - 1} (2\pi )^{1/z} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{z^2 }}} \right)} \right) \\ & = \frac{z}{{\rm e}} + \frac{{a - 1/2}}{{\rm e}}\log z + \frac{{\log (2\pi )}}{{\rm e}} + \mathcal{O}\!\left( {\frac{{\log z}}{z}} \right) \end{align*} as $z\to +\infty$. Hence, $$\tag{2} (\Gamma (z + a))^{1/z} - (\Gamma (z - 1 + a))^{1/(z - 1)} = \frac{1}{{\rm e}} + \mathcal{O}\!\left( {\frac{1}{z}} \right) $$ as $z\to +\infty$. From $(1)$ and $(2)$ it follows that $f(x)-f(x-1) \to 1/{\rm e}$.