# show that for any irrational $\alpha$ the limit $\lim _{n \to\infty} \sin n \alpha \pi$ does not exist .. [duplicate]

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

## marked as duplicate by dxiv, Xander Henderson, Daniel Fischer real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 6 '18 at 13:10

• 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 – user566930 Jul 5 '18 at 22:10

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 $|\sin n\alpha \pi|<0.5.$
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.