3
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ Observations suggest that the sum is equal to $(n-1)/2n$, but I am still seeking for a proof. $\endgroup$ – Sangchul Lee Aug 19 '14 at 10:33
5
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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 $1/\sqrt e$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.