Find the limit point of the sequence $\{s_n\}$ given by $s_n=\cos n $.
I know by this post Limit of sequence $s_n = \cos(n)$ that the sequence does not converge. But I don't know how to search those points.
Asked
Active
Viewed 5,162 times
5
-
4You'll be surprised to learn that the answer is $[-1,1]$ ! Can you think of a proof ? – Gabriel Romon Jun 22 '14 at 15:23
-
4There's one fact that might be unfamiliar to you but will help with the proof: the sequence of fractional parts $1/\pi-\lfloor1/\pi\rfloor$, $2/\pi-\lfloor2/\pi\rfloor$, ..., $n/\pi-\lfloor n/\pi\rfloor$, ... is dense in the interval $[0,1)$ - meaning that every tiny subinterval $(a,b)\in[0,1)$ contains infinitely many elements of the sequence. This follows from the fact that $1/\pi$ is irrational. – Greg Martin Jun 22 '14 at 15:51
-
Why this set is dense on $[0,1)$? – EQJ Jun 22 '14 at 16:03
-
@G.T.R after I have been thinking about this problem for a while but I get stuck. Can you give me a hint? – EQJ Jun 23 '14 at 03:41
1 Answers
7
The set $A=\{n+2\pi k:n,k\in\mathbb{Z}\}$ is dense on $\mathbb{R}$. Given a $y\in[-1,1]$ there existe an $x\in\mathbb{R}$ such that $\cos x=y$. Since $A$ is dense on $\mathbb{R}$ there existe a sequence $s_m=n_m+2\pi k_m$ of elements in $A$ such that $\lim\limits_{m\rightarrow\infty}s_m=x$. Then $$\lim\limits_{m\rightarrow\infty}\cos {n_m}=\lim\limits_{m\rightarrow\infty}\cos(n_m+2\pi k_m)=\cos \left(\lim\limits_{m\rightarrow\infty}(n_m+2\pi k_m)\right)=\cos x=y.$$ This implies that the limits points of $\cos n$ is all point in $[-1,1]$.
-
2The claim in the first sentence is (more or less) proved in an answer to the following question – Julian Kuelshammer Apr 30 '15 at 14:52