How prove this limit $\left(\frac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\lfloor\frac{mk}{n}\rfloor}\{\frac{mk}{n}\} \right)^n=\frac{1}{\sqrt{e}}$ let $m$ is even number,and $n$ is  odd number,and such $(m,n)=1$,
show this limit:
$$\lim_{n\to\infty}\left(\dfrac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\left\lfloor\dfrac{mk}{n}\right\rfloor}\left\{\dfrac{mk}{n}\right\} \right)^n=\dfrac{1}{\sqrt{e}}$$
where $\{x\}=x-\lfloor x\rfloor$
I think we can find this sum $$\sum_{k=1}^{n-1}(-1)^{\left\lfloor\dfrac{mk}{n}\right\rfloor}\left\{\dfrac{mk}{n}\right\}$$
But I can't
 A: Oh, I think I am late, but decided not to erase it.

Let $m = 2l$. We claim that the quantity
$$\lfloor mk / n \rfloor \equiv \lfloor 2lk / n \rfloor \pmod 2$$
depends only on $r_{k} = lk \text{ mod } n$. Indeed, write $lk = nq + r_{k}$. Then
$$ \lfloor 2lk / n \rfloor
= \lfloor 2(nq + r) / n \rfloor
= 2q + \lfloor 2r_{k} / n \rfloor. $$
So it follows that
$$ S_{n} := \sum_{k=1}^{n-1} (-1)^{\lfloor mk/n \rfloor} \left\{ \frac{mk}{n} \right\} = \sum_{k=1}^{n-1} (-1)^{\lfloor 2r_{k}/n \rfloor} \left\{ \frac{2r_{k}}{n} \right\} $$
But since $(l, n) = 1$, $r_{k}$ is just a rearrangement of $1, \cdots, n-1$ and we have
$$ S_{n} := \sum_{k=1}^{n-1} (-1)^{\lfloor 2k/n \rfloor} \left\{ \frac{2k}{n} \right\}. $$
Dividing the sum into two parts, with one running over $k = 1, \cdots, \frac{n-1}{2}$ and the other running over $k = \frac{n+1}{2}, \cdots, n-1$, it follows that
$$ S_{n} = \sum_{k=1}^{(n-1)/2} \frac{2k}{n} - \sum_{k=(n+1)/2}^{n-1} \left( \frac{2k}{n} - 1 \right) = \frac{n-1}{2n}. $$
This proves the observation as desired.

A slight generalization: Let $f$ be a $C^{2}$-function on $[0, 1]$. Then utilizing the Taylor's Theorem, it is not hard to show that
\begin{align*}
S_{f,n} := \sum_{k=1}^{n-1} (-1)^{\left\lfloor \frac{mk}{n} \right\rfloor} f \left( \left\{ \tfrac{mk}{n} \right\} \right)
&= \sum_{k=1}^{(n-1)/2} \left( f \left( \tfrac{2k}{n} \right) - f \left( \tfrac{2k}{n} - \tfrac{1}{n} \right) \right) \\
&= \frac{1}{2}(f(1) - f(0)) - \frac{1}{4n}(f'(1) + f'(0) + o(1)).
\end{align*}
So we have
$$ \left( 1 - \frac{f(1) - f(0)}{2} + S_{f,n} \right)^{n} \to \exp\left(-\frac{f'(0)+f'(1)}{4} \right). $$
A: The observation of @sos440 is true. Here's a proof. 
Write $mk=nq_k + r_k$, with $0<r_k\leq n-1$. 
Then we see that parity of $q_k$ and $r_k$ are equal. 
Thus the sum in question after rearranging equals 
$$
\sum_{r=1}^{n-1} (-1)^r \frac{r}{n}.$$
(Here we use $(m,n)=1$. )
This indeed equals 
$$\frac{n-1}{2n}.$$
Therefore the limit in question is $1/\sqrt e$. 
