Let $n\in\mathbb{N}$.

From the book *"Uniform Distribution of Sequences"* (available [here](https://web.maths.unsw.edu.au/~josefdick/preprints/KuipersNied_book.pdf)) by *L. Kuipers* and *H. Niederreiter*, (from pg. 8) I found that for any irrational $\theta$, the sequence $(n\theta)$, $n = 1, 2, 3,... ,$ is uniformly distributed (or equidistributed) mod $1$ in the interval $(0,1)$, which is, in fact, Equidistribution theorem. This means that the sequence {$\pi n$}, $n=1,2,3,...,$ is equidistributed in the interval $(0,1)$, where {.} denotes the fractional part function. 

It can be easily proved by **Weyl's criterion** which states that the sequence $a_n$ is equidistributed modulo $1$ if and only if for all non-zero integers h,
$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}e^{2\pi ih(a_k)}=0$$
If we fix $a_n=\pi n$, it is easy to prove above criterion (proved [here](https://en.wikipedia.org/wiki/Equidistributed_sequence#Example_of_usage)). 

A thought came in my mind that if we fix $a_n=\pi (2n+1)!$ , which is a subsequence of previous sequence, is *Weyl's criterion* satisfied **?**
 In other words, is the sequence ($\pi (2n+1)!$), $n=1,2,...,$ equidistributed mod $1$ in the interval $(0,1)$ ? i.e, for all non-zero integers h, is the following true?

$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}e^{2\pi ih\pi (2k+1)!}=0$$

I **conjecture** that the sequence {$\pi (2n+1)!$} is equidistributed in the interval $(0,1)$. This conjecture is equivalent to the affirmative answer of above two questions. 

I also conjecture a **weaker** statement which is as follows:

**"**The sequence {$\pi (2n+1)!$}, $n=1,2,3,...,$ will never approach 1.**"**

Can anyone please Prove/Disprove my conjecture? Proving/disproving my weaker conjecture will also be appreciated.