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1)show that for any irrational $\alpha $ the limit $\lim _{n\to\infty} \sin n \alpha \pi $ does not exist ..

2) show that for any rational $\alpha $ the limit $\lim _{n \rightarrow \infty} \sin (n! \alpha \pi) $ exist ?

My attempts : For 1) I'm very confused ?????

For 2) let $ \alpha =\frac {p}{q}$ with $p \in \mathbb{Z}$ and $q \in \mathbb{N}$ . For $n > q $ the number $n ! \alpha \pi$ is a multiple of $\pi $, which means that the terms of the sequence , beginning with some Value $ n_{0} $of the index n, are all equal to $0$

as in 1) I'm very confused. How can I approach this kind of problems ??

Pliz help me

Thanks un

Bernard
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jasmine
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    If you can get approximations $|\alpha-p/n|<1/n^2$ for many $n$'s, then $|n\alpha\pi-p\pi|<\pi/n$, which means that $\sin(n\alpha \pi)$ is arbitrarily close to $\sin(p\pi)=0$. Likewise, if you can get approximations $|2\alpha-p/n|<1/n^2$ for many $n$'s with $p$ odd, then $|n\alpha\pi-p\pi/2|<\pi/(2n)$. This means that $\sin(n\alpha\pi)$ is very close to $\sin(p\pi/2)=\pm1$. Now look at some of what is known about rational approximation –  Jul 05 '18 at 22:10

2 Answers2

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Hint: Let $\alpha$ be irrational. Prove that $$ \mathbb N \cdot \alpha \mod 2 $$ is dense in $[0,2]$. Hence $$ (\mathbb N \cdot \alpha \cdot \pi) \mod 2 \pi $$ is dense in $[0,2\pi]$. By the continuity of $\sin$ the result follows.

Stefan Mesken
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For 1).

For real $x$ let $[x]$ be the largest integer not exceeding $x.$ Let $\{x\}$ denote the non-negative fractional part of $x.$ That is, $\{x\}=x-[x].$

It is an old result, provable by elementary means, that if $\alpha$ is an irrational real then $\{\{n\alpha\}: n\in \Bbb N\}$ is dense in $[0,1].$

So there are infinitely many $n\in \Bbb N$ such that $1/3<\{n\alpha\}<1/2$ and there are infinitely many $n\in \Bbb N $ such that $ 0<\{n\alpha\}<1/6.$

So there are infinitely many $n\in \Bbb N$ such that $|\sin n\alpha \pi|>\sqrt 3\;/2> 0.866$ and there are infinitely many $n\in \Bbb N$ such that $|\sin n\alpha \pi|<0.5.$

DanielWainfleet
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