Showing $a_n=\sin(n)$ does not converge [duplicate]

Solution 1:

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.

Solution 2:

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 $[-1,1]$.