What is $\lim\limits_{x\to \infty}[g(x)-g(x-1)]\overset?=$

Question

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 .

Answer 1

$$g\left(x\right)=\int_{0}^{x}\prod_{n=2}^x\left(\frac{\log\left(y+2n\right)}{\log\left(y+2n-1\right)}\right)^{\log\left(n\right)}dy$$

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 $(R^2=0.999994)$ $$ g(x)-g(x-1)=a -\frac b{k+c}$$

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & \color{red}{ 3.21419} & 0.03247 & \{3.11085,3.31754\} \\ b & 3.93983 & 0.27615 & \{3.06099,4.81868\} \\ c & 1.55770 & 0.13252 & \{1.13597,1.97943\} \\ \end{array}$$

Answer 2

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 $2\sqrt{e}$