4

Show that $a_n=\sin(n)$ does not converge

My idea:

Take two subsequences: $a_{n_k}=\sin(\frac {\pi k} 2)$ , $a_{n_l}=\sin(\frac {2\pi l} 3)$

So: $\forall n$ : $\lim_{n\to\infty} a_{n_k}=1$, $\lim_{n\to\infty} a_{n_l}=-1$

So two inifinite subsequence converge to different limits thus the sequence $a_n$ doesn't converge.

Is that correct ?

Edit:

The contrapositive of the defintion of a limit of a sequence:

$\exists\epsilon>0 : \forall n\in N : \exists n>N \Rightarrow |x_n-L|>\epsilon$

Take $\epsilon =1$ and we know that $\sin(n)$ is bounded so lets take it's supermum: 1

$|1-1|>\epsilon=1\Rightarrow 0>1 \Rightarrow$ Contradiction.

Caleb Stanford
  • 45,895
GinKin
  • 4,429
  • 1
    The idea is correct, the writing however... – Git Gud Apr 07 '14 at 10:26
  • The writing is terrible. How can $a_{n_k}$ depend on $n$, not on $k$? – 5xum Apr 07 '14 at 10:26
  • 4
    $a_n$ is defined, I assume on natural numbers $n\in\mathbb{N}$, so you can't define a subsequence $a_{n_k}$ unless $n_k\in\mathbb{N}$. Instead, recall the definition of convergence, and try to write the contrapositive of that definition here. I'll see if you get it correct - that should help more. – Maylor Apr 07 '14 at 10:27
  • @5xum alright alright, now it depends on n. – GinKin Apr 07 '14 at 10:31
  • @Maylor I thought it's obvious that indexes of a sequence are natural numbers. – GinKin Apr 07 '14 at 10:32
  • 1
    @GinKin Yes, it depends on $n$, instead of on $k$. I still have no idea what $a_{n_k}$ is for $k=121$, for example. – 5xum Apr 07 '14 at 10:45
  • @5xum okay, I think I got what you mean now. – GinKin Apr 07 '14 at 10:49
  • No, it's still bad. See, there is no value of $n$ for which $a_n$ equals $\frac\pi2$ so what is written now is still wrong. What I think you want is to define $n_k = \frac{\pi k}{2}$ – 5xum Apr 07 '14 at 10:51
  • @Maylor, I tried to use the contrapostive with the limit definition. Can you have a look please ? – GinKin Apr 07 '14 at 10:53
  • @5xum but the index is a natural number... – GinKin Apr 07 '14 at 10:55
  • @GinKin Which is why I think your whole argument is wrong. Look, it's simple. If you have a sequence ${a_n}{n=1}^{\infty}$, you can define a subsequence of $a_n$ by selecting some values of $n$, namely $n_1,n_2,\dots$. THAT gives you a sequence $${a{n_k}}_{k=1}^{\infty}.$$ In your case, you say $a_n=\frac{\pi n}{2}$, which makes no sense. – 5xum Apr 07 '14 at 10:57
  • @5xum I forgot to input the $\sin()$ and didn't notice it... Does it make sense now ? – GinKin Apr 07 '14 at 11:09
  • 1
    @GinKin Since a sequence is convergent if and only if it is Cauchy, we can just try and show that the sequence is not Cauchy. The contrapositive is as follows: $\exists\epsilon > 0$ such that $\forall N\in\mathbb{N}$, $\exists m, n>N$ such that $|sin(m)-sin(n)|>\epsilon$. – Maylor Apr 07 '14 at 11:52

2 Answers2

18

Your argument is not correct since you compute values $\sin{k\pi\over 2}$ therein, which are not values of the original sequence.

Here is a proof that does not make use of density arguments:

Assume $\lim_{n\to\infty}\sin n=\sigma\in{\mathbb R}$. Then $$2\cos n\>\sin 1=\sin(n+1)-\sin(n-1)\to 0\qquad(n\to\infty)\ ,$$ which implies $\lim_{n\to\infty}\cos n=0$, whence $\sigma\in\{-1,1\}$. Letting $n\to\infty$ in $$\sin(n+1)=\sin n\>\cos 1+\cos n\>\sin 1$$ would then imply $\cos 1=1$, which is clearly wrong.

Christian Blatter
  • 226,825
4

This needs a deeper result. Let the fractional part of $x$ be denoted by $\{x\}$. By Kronecker's Theorem, the set $\left\{ \{\frac{n}{2\pi}\}:n\in\Bbb{N}\right\}$ is a dense subset of $[0,1]$, (because $\pi$ is irrational.) So, there exist two sequences of integers $(n_k)_k$ and $(m_k)_k$ such that

$$ \lim_{k\to\infty}\left\{\frac{n_k}{2\pi}\right\}=0,\quad\hbox{and}\quad\lim_{k\to\infty}\left\{\frac{m_k}{2\pi}\right\}=\frac{1}{4} $$ Equivalently $$ \lim_{k\to\infty}\left(n_k-2\pi\left\lfloor\frac{n_k}{2\pi}\right\rfloor\right)=0,\quad\hbox{and}\quad\lim_{k\to\infty} \left(m_k-2\pi\left\lfloor\frac{m_k}{2\pi}\right\rfloor\right)=\frac{\pi}{2} $$ That is $\lim\limits_{k\to\infty}\sin(n_k)=0$ and $\lim\limits_{k\to\infty}\sin(m_k)=1$. Thus the sequence $(\sin(n))_n$ does not converge.

$\bf{Remark.}$ A variation on this proof shows that the set $\{\sin n:n\in\Bbb{N}\}$ is dense in the interval $[-1,1]$.

Omran Kouba
  • 28,772