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

From the book "Uniform Distribution of Sequences" (available here) 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).

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.

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

From the book "Uniform Distribution of Sequences" (available here) 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).

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." i.e $$\lim_{n \to \infty}\{\pi (2n+1)!\}≠1$$

where {.} is the fractional part function. Can anyone please Prove/Disprove my conjecture? Proving/disproving my weaker conjecture will also be appreciated.

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

From the book "Uniform Distribution of Sequences" (available here) 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).

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:

{$π(2n+1)!$}$≤0.5$ infinitely often, which is as same as $\sin(2π^2(2n+1)!)>0$ infinitely often (Graph available here).

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

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

From the book "Uniform Distribution of Sequences" (available here) 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).

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.

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

From the book "Uniform Distribution of Sequences" (available here) 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).

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 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:

{$π(2n+1)!$}$≤0.5$ infinitely often, which is as same as $\sin(2π^2(2n+1)!)>0$ infinitely often (Graph available here).

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

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

From the book "Uniform Distribution of Sequences" (available here) 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).

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:

{$π(2n+1)!$}$≤0.5$ infinitely often, which is as same as $\sin(2π^2(2n+1)!)>0$ infinitely often (Graph available here).

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