I solved the problem. But I have some doubts on the solution.

Question

Given the sequence $a_n = \sin(n)$, where $n$ is an integer. Prove that limit superior is $1$.

My solution: by previous theorems we know that if $x,y \in \Bbb R$ and the ratio of $x$ and y is irrational, then there exists the integers $a$ and $b$ such that $$\forall \epsilon, \quad 0<ax+by<\epsilon$$ So we can put integrs $m$ ands $n$ such that $$0<\frac{n}{m}-\frac{\pi}{2}<\epsilon$$ And we are done.

But here m should be in form of $m=4k+1,\, k\in \Bbb Z$.

My question is, how we can be sure that we can find such $m$?

Answer 1

What you actually need is, given $\epsilon>0$, to show there exist integers $n,k$ such that $$| (4k+1)\pi/2-n |= |2k\pi+\pi/2-n|<\epsilon \,.$$ This follows from the density of the orbit of irrational rotations https://en.wikipedia.org/wiki/Irrational_rotation

Prove that the orbit of an iterated rotation of 0 (by (A)(Pi), A irrational) around a circle centered at the origin is dense in the circle.

Irrational angle rotation argument.

Dense set in the unit circle- reference needed

Answer 2

Let $x$ and $y$ be real numbers. And suppose that $ \frac{x}{y}$ is irrational.

Prove the following: for each $\varepsilon>0$, there exist integers a and b such that $0<ax+by< \varepsilon$.

Consider the set $S=${$ax+by>0$: $x,y \in R$, $ \frac{x}{y}$ is irrational, $a,b \in Z$}.

First to show $d= \inf S \ge 0$ exists we must show $S$ is non-empty and $0$ is lower bound of $S$. The set $S$ is obviously non-empty. Just consider $|x|+|y|=±1⋅x+(±1)⋅y>0$. And by definition, if $ax+by∈S$ then $ax+by>0$ so $0$ is a lower bound.

Assume that $c=min$ $S$. Then I claim that $x$ and $y$ are both integer multiples of $c$. $x=cm$, $y=nm$ and $n,m \in Z$.

If not then $x=cm+r$ where $0\leq r<c$ and $ m \in Z$. From here we have $r=x - mc= x - m(kx+ty)=(1-mk)x+(-tm) y$ and $k,t \in Z$. So $r \in S$ and $r<c$. It contradicts with the fact that $c =\min S$. Similarly we can show $y=cn$ for some $n \in Z$.

But $ \frac{x}{y} = \frac{m}{n}$ sould be irrational where $ m,n \in Z$ which is immposible. It means that the set $S$ has no minimum.

Now..... Suppose $d=\inf S≠0$. Then $\inf S>0$. That mean $\frac{3}{2}d>d$ so $2d>d$ is not a lower bound of $S$. We also know $d∈S$ is impossible as that would mean $d=\min S$ and we know $\min S$ doesn't exist.

So there exists $ax+by$ so that $d<ax+by<2d$. And as $ax+by>d$, $ax+by$ can't be a lower bound so there must exist $ex+fy$ so that $d<ex+fy<ax+by<2d$.

So $0<(a−e)x+(b−f)y<d=\inf S$ which is impossible as $0<(a−e)x+(b−f)y⟹(a−e)x+(b−f)y∈S$.

So $d=infS=0$.

.......

Which means for any $ϵ>0$, then $ϵ$ is not a lower bound of $S$ so there is an $ax+by∈S$ so that $0<ax+by<ϵ$.

Answer 3

if you want From the theorem above You can deduce that you can find infinitely pairs of $n,k \in Z$ such that $0< \frac{\pi}{2}+2* \pi*k-n< \varepsilon$. And dont worry about form of the denominator like $ \frac{\pi}{2} \approx \frac{n}{4k+1}$ becasuse actually what you need is the ratio and with infinitely many n, k you are able form any kind of ratio. Because in $Q$ class of ratio infinitely many. for example $\frac{1}{2}$=${\frac{1}{2}, \frac{2}{4}, \frac{4}{8} .....}$