Showing that for $x > e^{2.5102}, 0 \le \lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506x}{\ln x}\rfloor \le 1$ Showing that for $x > e^{2.5102}, 0 \le \lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506x}{\ln x}\rfloor \le 1$
Does this argument work:
(1)  $\dfrac{1.25506x}{\ln x}$ is increasing for $x > e$ with since:


*

*$\dfrac{d}{dx}(\dfrac{1.25506x}{\ln x}) = \dfrac{1.25506\ln(x)-1.25506}{(\ln x)^2}$

*$1.25506\ln(x) - 1.25506$ is positive
(2)  For $x > e^{2.51012}$, since $\dfrac{1.25506x}{\ln x}$ is increasing for $ x > 1$:
$$0 < \dfrac{1.25506(x+1)}{\ln(x+1)} - \dfrac{1.25506x}{\ln x} < \dfrac{1.25506(x+1)}{\ln x} - \dfrac{1.25506x}{\ln x} = \dfrac{1.25506}{\ln x} < 0.5$$
(3) There exists integers $a,b$ such that $0 \le a < 1$ and $0 \le b < 1$ such that
$$\lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506(x)}{\ln(x)}\rfloor = \dfrac{1.25506(x+1)}{\ln(x+1)} - a - \dfrac{1.25506(x)}{\ln(x)} + b$$ 
(4)  Since $-1 < b - a < 1$, it follows that:
$$-1 < \lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\rfloor - \lfloor\dfrac{1.25506(x)}{\ln(x)}\rfloor < 0.5 + 1 = 1.5$$
Does the conclusion follow?  Did I make a mistake?  Is there a better way to establish the same conclusion?

(1)(2) are correct though you have a typo in (2). It should be $x\gt e$ instead of $x\gt 1$.
(3) is not correct. If $a,b$ are integers such that $0\le a\lt 1$ and $0\le b\lt 1$, then $a=b=0$ for which
$$\left\lfloor\dfrac{1.25506(x+1)}{\ln(x+1)}\right\rfloor - \left\lfloor\dfrac{1.25506(x)}{\ln(x)}\right\rfloor = \dfrac{1.25506(x+1)}{\ln(x+1)} - a - \dfrac{1.25506(x)}{\ln(x)} + b$$ 
does not necessarily hold since the RHS is not necessarily an integer.

Let $f(x):=\dfrac{1.25506(x)}{\ln(x)}$.
After getting $0\lt f(x+1)-f(x)\lt 0.5$ in (2), we can separate it into two cases :


*

*If there exists an integer $N$ such that $f(x)\lt N\le f(x+1)$, then $\lfloor f(x)\rfloor=N-1$ and $\lfloor f(x+1)\rfloor=N$ so $\lfloor f(x+1)\rfloor-\lfloor f(x)\rfloor=N-(N-1)=1$.

*If there exists an integer $N$ such that $N\le f(x)\lt f(x+1)\lt N+1$, then $\lfloor f(x)\rfloor=\lfloor f(x+1)\rfloor=N$ so $\lfloor f(x+1)\rfloor-\lfloor f(x)\rfloor=N-N=0$.
Therefore, $0\le \lfloor f(x+1)\rfloor -\lfloor f(x)\rfloor\le 1$ follows.

Added : 
If you meant $a,b$ are real numbers, not integers, then what you did is correct.
In conclusion, you didn't make any mistakes. You just had a few typos.
