# A question involving e, floor, and all x > 0

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 $x$.)

• If you mean for all positive numbers $x>0$ (not just integers), then this surely fails, because the two sides jump at different values. – GH from MO Jun 2 '15 at 18:55
• E.g. $x = 0.62$ is a counterexample -- there the left-hand side is $3$, while the right-hand side is $2$. – Stefan Kohl Jun 2 '15 at 18:57
• It is, though, true for all integers $n \leq 10^7$ (gp in under 2 minutes). – Noam D. Elkies Jun 2 '15 at 19:06
• I take it $x+1/2$ means $x+(1/2)$, and not $(x+1)/2$. – Gerry Myerson Jun 2 '15 at 23:31

## 1 Answer

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.

• I proved a very similar inequality over at math.SE, taking a similar approach. math.stackexchange.com/questions/921880 I haven't quite unpacked the notation enough to see if this is the same question. – David E Speyer Jun 3 '15 at 2:36
• Okay, I checked, it is the same question. – David E Speyer Jun 3 '15 at 2:54