$\lim\limits_{x\to \infty}[f(x)-f(x-1)]\overset{?}{=}e$ 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 :$\lim_{x\to\infty}f(x)-f(x-1)=\frac{egg}{gg}$
 A: Here is a weaker result which is still enough to show that the limit is not $e$:

Claim. We have
$$ \lim_{n\to\infty} \frac{f(n)}{n} = \sqrt{3} + 2\log(1+\sqrt{3}) - \log 2 \approx 3.04901 $$

Proof. Let $n$ be a positive integer. Then
$$ f(n) = \int_{0}^{n} \prod_{j=1}^{n} \frac{\log(y+2j-1)}{\log(y+2j)} \frac{y+2j}{y+2j-1} \, \mathrm{d}y. $$
Substituting $y = nt$, the integral is recast as
\begin{align*}
f(n) 
&= n \int_{0}^{1} g_n(t) \, \mathrm{d}t,
\qquad g_n(t) := \prod_{j=1}^{n} \frac{\log(nt+2j-1)}{\log(nt+2j)} \frac{nt+2j}{nt+2j-1}. \tag{*}
\end{align*}
Using the inequality $\frac{1}{1-x} \leq \exp(x + x^2) $ for $x \in [0, \frac{1}{2}]$, we find that
\begin{align*}
g_n(t)
&= \prod_{j=1}^{n} \frac{\log(nt+2j-1)}{\log(nt+2j)} \frac{1}{1 - \frac{1}{nt+2j}} \\
&\leq \exp\left[ \sum_{j=1}^{n} \left( \frac{1}{nt+2j} + \frac{1}{(nt+2j)^2} \right) \right] \\
&\leq \exp\left[ \int_{0}^{n} \frac{1}{nt+2s} \, \mathrm{d}s + \sum_{j=1}^{\infty} \frac{1}{(2j)^2} \right] \\
&= \exp\left[ \frac{1}{2} \log \left(\frac{t+2}{t}\right) + \frac{\zeta(2)}{4} \right].
\end{align*}
This proves that $g_n(t) \leq Ct^{-1/2}$ uniformly in $n$, and so, we can apply the dominated convergence theorem provided $g_n(t)$ converges pointwise as $n \to \infty$. However, for each fixed $t \in (0, 1]$,
\begin{align*}
g_n(t)
&= \prod_{j=1}^{n} \biggl( 1 + \frac{\log(1 - \frac{1}{nt + 2j})}{\log(nt+2j)} \biggr) \frac{1}{1 - \frac{1}{nt+2j}} \\
&= \exp \left[ \sum_{j=1}^{n} \biggl( \frac{1}{nt + 2j} + \mathcal{O}\biggl( \frac{1}{n \log n} \biggr) \biggr) \right] \\
&= \exp \left[ \sum_{j=1}^{n} \frac{1}{t + 2(j/n)} \frac{1}{n} + \mathcal{O}\left( \frac{1}{\log n} \right) \right] \\
&\to \exp\left( \int_{0}^{1} \frac{\mathrm{d}s}{t + 2s} \right)
= \sqrt{\frac{t + 2}{t}}.
\end{align*}
Therefore, by the dominated convergence theorem,
$$ \lim_{n\to\infty} \frac{f(n)}{n}
= \int_{0}^{1} \sqrt{\frac{t + 2}{t}} \, \mathrm{d}t
= \boxed{\sqrt{3} + 2\log(1+\sqrt{3}) - \log 2}. $$
A: 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 $1+\sqrt{e}$.
