Is $\lfloor(x+1/2)e\rfloor = \lfloor(x+1)(1+1/x)^x\rfloor$ for all $x > 0$?


The question occurred in connection with (nonhomogeneous) Beatty sequences, $\lfloor nr+h\rfloor$, where irrational $r>0$ and real $h$ are fixed, and $n = 1,2,\dots$.


Let


$$s\_n = (n+1/2)e - (n+1)(1+1/n)^n$$ I checked that $(s\_n)$ is strictly decreasing and $0 < s\_n < 1$ for $n = 1,2,\dots, 10^6$.


Oops, thanks for noting that the leap from positive integers $n$ to real $x$ was blind. So, the answer to the question as asked is "no" - leaving a subquestion, whether the proposed identity holds for positive integers $n$. (Still, though, should anything else be said about the left side versus the right side for non-integer values of $x$.) 



A question involving e, floor, and all x > 0
The question is whether there is an integer $m$ with $(n+1/2)e < m \le (n+1)(1+1/n)^n$. Note that $s\_n \sim e/n$, so this would mean (approximately)
$$ 0 > e - \dfrac{2m}{2n+1} > \dfrac{2e}{n(2n+1)}$$


Now the continued fraction for $e$ is well-known:


$$ e = [2;1,2,1,1,4,1,1,6,1,1,8,\ldots]$$


The corresponding even-numbered convergents ($[2;1],\; [2;1,2,1],\; \ldots$) are greater than $e$, the odd-numbered ones are less than $e$. But the even-numbered convergents all have odd numerators. So we won't get a counterexample from those. I suspect that with further work on the "not-quite-best" rational approximations of $e$ one can show that there are no counterexamples.


