$\lim\limits_{x\to \infty}[f(x)-f(x-1)]\overset{?}{=}e$

Question

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 :$\lim_{x\to\infty}f(x)-f(x-1)=\frac{egg}{gg}$

Answer 1

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,

$$ \lim_{n\to\infty} \frac{f(n)}{n} = \int_{0}^{1} \sqrt{\frac{t + 2}{t}} \, \mathrm{d}t = \boxed{\sqrt{3} + 2\log(1+\sqrt{3}) - \log 2}. $$

Answer 2

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 $1+\sqrt{e}$.

Answer 3

Too long for a comment

Using $x=10^k$ and computing $$\left( \begin{array}{cc} k & f(x)-f(x-1) \\ 1 & 1.80609 \\ 2 & 2.25907 \\ 3 & 2.48912 \\ 4 & 2.62061 \\ 5 & 2.70309 \\ 6 & 2.75896 \\ \end{array} \right)$$

A quick and dirty nonlinear regression $(R^2=0.999998)$ $$ f(x)-f(x-1)=a -\frac b{k+c}$$

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & \color{red}{3.12067} & 0.01415 & \{3.07564,3.16571\} \\ b & 2.45391 & 0.09706 & \{2.14503,2.76278\} \\ c & 0.86524 & 0.05706 & \{0.68365,1.04682\} \\ \end{array}$$

Computing $$g\left(x\right)=\int_{0}^{x}\prod_{n=1}^{x}\frac{\left(y+2n\right)}{\left(y+2n-1\right)}dy=\int_{0}^{x} \frac{\Gamma \left(\frac{y+1}{2}\right) \Gamma \left(x+\frac{y+2}{2}\right)}{\Gamma \left(\frac{y+2}{2}\right) \Gamma \left(x+\frac{y+1}{2}\right)}\,dy$$ $$\left( \begin{array}{cc} k & g(x)-g(x-1) \\ 1 & 2.70835 \\ 2 & 2.94330 \\ 3 & 3.01564 \\ 4 & 3.03846 \\ 5 & 3.04568 \\ 6 & 3.04841 \\ \end{array} \right)$$

Another quick and dirty nonlinear regression $(R^2=0.999996)$ $$ g(x)-g(x-1)=a -\frac b{k-c}$$

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & \color{red}{3.09874} & 0.01141 & \{3.06245,3.13504\} \\ b & 0.24011 & 0.04133 & \{0.10857,0.37165\} \\ c & 0.38563 & 0.09176 & \{0.09359,0.67766\} \\ \end{array}$$