# 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

• Observations suggest that the sum is equal to $(n-1)/2n$, but I am still seeking for a proof. – Sangchul Lee Aug 19 '14 at 10:33

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

$$\left( 1 - \frac{f(1) - f(0)}{2} + S_{f,n} \right)^{n} \to \exp\left(-\frac{f'(0)+f'(1)}{4} \right).$$

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 $1/\sqrt e$.