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.

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.