$\lim\limits_{x\to \infty}[f(x)-f(x-1)]\overset{?}{=}e$ Let :
$$f\left(x\right)=\int_{0}^{\lfloor x\rfloor}\prod_{n=1}^{\lfloor x\rfloor}\frac{\left(y+2n\right)\ln\left(y+2n-1\right)}{\left(y+2n-1\right)\ln\left(y+2n\right)}dy$$
Conjecture:
$$\lim_{x\to \infty}f(x)-f(x-1)=e$$
We have for $x=150000$:
$$f(x)-f(x-1)\approx 2.714446$$
I cannot proceed further because Desmos is a bit unfriendly.
How i come up with this conjecture :
Have a look to Andersson's inequality https://www.sciencedirect.com/science/article/pii/S0893965905003666 :
$$\int_{0}^{1}f_{1}\left(x\right)f_{2}\left(x\right)...f_{n}\left(x\right)dx\ge\frac{2n}{n+1}\left(\int_{0}^{1}f_{1}\left(x\right)dx\right)\left(\int_{0}^{1}f_{2}\left(x\right)dx\right)...\left(\int_{0}^{1}f_{n}\left(x\right)dx\right)$$
here :
$$f_{n}\left(x\right)=\frac{\left(x+2n\right)\ln\left(x+2n-1\right)}{\left(x+2n-1\right)\ln\left(x+2n\right)}$$
It doesn't fullfilled the constraint $f_n(0)=0$ and the convexity for $n$ small on $x\in[0,1]$ .
On the other hand the use of the floor function is just a pratical graph point of view on Desmos (free software) enlightening the probable existence of an asymptote .
Does it converge? If yes, is it $e$?
A counter-example is also welcome!
Ps : Something weird should be :$\lim_{x\to\infty}f(x)-f(x-1)=\frac{egg}{gg}$

I show that the limit is finite :
We have :
$$f\left(x\right)=\int_{0}^{\lfloor x\rfloor}\prod_{n=1}^{\lfloor x\rfloor}\frac{\left(y+2n\right)\ln\left(y+2n-1\right)}{\left(y+2n-1\right)\ln\left(y+2n\right)}dy<\int_{0}^{\lfloor x\rfloor}\prod_{n=1}^{\lfloor x\rfloor}\frac{\left(y+2n\right)}{\left(y+2n-1\right)}dy$$
Using Am-Gm :
$$\int_{\lfloor x\rfloor}^{\lfloor x+1\rfloor}\prod_{n=1}^{\lfloor x\rfloor}\frac{\left(y+2n\right)\ln\left(y+2n-1\right)}{\left(y+2n-1\right)\ln\left(y+2n\right)}dy+\int_{0}^{\operatorname{floor}\left(x\right)}\frac{1}{n}\sum_{k=1}^{1+n}\left(\frac{\left(y+2k\right)}{\left(y+2k-1\right)}\right)^{n}-\frac{1}{n}\sum_{k=1}^{n}\left(\frac{\left(y+2k\right)}{\left(y+2k-1\right)}\right)^{n}dy\simeq f(x+1)-f(x) $$
This telescoping and have as value $0$ for :
$$\int_{0}^{\operatorname{floor}\left(x\right)}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{\left(y+2k\right)}{\left(y+2k-1\right)}\right)^{n}-\frac{1}{n}\sum_{k=1}^{n}\left(\frac{\left(y+2k\right)}{\left(y+2k-1\right)}\right)^{n}dy$$
As $n\to \infty$
The last integral divided n is equal to :
$$n\sqrt{e}$$
We can conclude that this is finite is approximatively $1+\sqrt{e}$.
