What is $\lim\limits_{x\to \infty}[g(x)-g(x-1)]\overset?=$ Hi it's a follow up of $\lim\limits_{x\to \infty}[f(x)-f(x-1)]\overset{?}{=}e$
Let :
$$g\left(x\right)=\int_{0}^{\operatorname{floor}\left(x\right)}\prod_{n=2}^{\operatorname{floor}\left(x\right)}\left(\frac{\ln\left(y+2n\right)}{\ln\left(y+2n-1\right)}\right)^{\ln\left(n\right)}dy$$

What is :
$$\lim_{x\to \infty}\big[g(x)-g(x-1)\big]\overset?=$$

As we have  for $x>2$ and $n\geq 2$ an integer :
$$\frac{x+2n}{x+2n-1}-\left(\frac{\ln\left(x+2n\right)}{\ln\left(x+2n-1\right)}\right)^{\ln\left(n\right)}>0$$
I think the limit is less than the result due to user @SangchulLee ( see the linked question above )
The result is close to $1+\sqrt{e}$ and perhaps it involves hypergeometric function .
So what is the limit of the difference $\lim\limits_{x\to \infty}\big[g(x)-g(x-1)\big]\overset?=$
Sides notes
As in my other question we can apply Andersson's inequality :
With :
$$-h_k(x)=\frac{\left(xk\right)\left(\ln\left(x+2k\right)\right)^{\ln\left(k\right)}}{\left(xk+1\right)\left(\ln\left(x+2k-1\right)\right)^{\ln\left(k\right)}}$$
Where $k$ sufficiently large .
 A: $$g\left(x\right)=\int_{0}^{x}\prod_{n=2}^x\left(\frac{\log\left(y+2n\right)}{\log\left(y+2n-1\right)}\right)^{\log\left(n\right)}dy$$
Using as in the previous post $x=10^k$
$$\left(
\begin{array}{cc}
 k & g(x)-g(x-1)\\
 1 & 1.67606 \\
 2 & 2.09772 \\
 3 & 2.35404 \\
 4 & 2.51266 \\
 5 & 2.61550 \\
 6 & 2.68595 & \color{red}{\large >~1+\sqrt e}  \\
\end{array}
\right)$$
One more quick and dirty nonlinear regression $(R^2=0.999994)$
$$ g(x)-g(x-1)=a -\frac b{k+c}$$
$$\begin{array}{l|lll}
 \text{} & \text{Estimate} & \text{Std Error} &
   \text{Confidence Interval} \\
\hline
 a & \color{red}{ 3.21419} & 0.03247 & \{3.11085,3.31754\} \\
 b & 3.93983 & 0.27615 & \{3.06099,4.81868\} \\
 c & 1.55770 & 0.13252 & \{1.13597,1.97943\} \\
\end{array}$$
A: As $n\to \infty$:
$$\left(\frac{\ln\left(x+2n\right)}{\ln\left(x+2n-1\right)}\right)^{\ln(n)}\to 1$$
Using Am-Gm :
$$\int_{\lfloor x\rfloor}^{\lfloor x+1\rfloor}\prod_{n=2}^{\lfloor x\rfloor}\left(\frac{\ln\left(y+2n\right)}{\ln\left(y+2n-1\right)}\right)^{\ln(n)}dy+\int_{0}^{\operatorname{floor}\left(x\right)}\frac{1}{n+1}\sum_{k=2}^{1+n}\left(\frac{\ln\left(y+2k\right)}{\ln\left(y+2k-1\right)}\right)^{(n+1)\ln(n)}-\frac{1}{n}\sum_{k=2}^{n}\left(\frac{\ln\left(y+2k\right)}{\ln\left(y+2k-1\right)}\right)^{n\ln(n)}dy\simeq g(x+1)-g(x) $$
Now as $n\to \infty$:
$$\left(\frac{\ln\left(x+2n\right)}{\ln\left(x+2n-1\right)}\right)^{(n+1)\ln(n)}-\left(\frac{\ln\left(x+2n-2\right)}{\ln\left(x+2n-3\right)}\right)^{n\ln(n)}\to 0$$
Now I think we can do that using Fatou's lemma so we invert the limit but :
$$\left(\frac{\ln\left(x+2n\right)}{\ln\left(x+2n-1\right)}\right)^{(n+1)\ln(n)}\to \sqrt{e}$$
Wich gives as lower bound :
$$1+\sqrt{e}$$
The limit describes by Claude Leibovici seems close to $2\sqrt{e}$
