TITLE: A question involving e, floor, and all x > 0
QUESTION [9 upvotes]: 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 $x$.)

REPLY [6 votes]: 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.