Divergence of the sequence $\sin(n)$ 
Possible Duplicate:
Prove the divergence of the sequence $(\sin(n))_{n=1}^\infty$. 

How can I show that the sequence
$$
 a_n = \sin(n)
$$
is divergent? I tried to show that $\sin(n+1) - \sin(n)$ get always larger than some constant, but I did not succeeded.

Your $(a_n)$ is bounded, but it doesn't converge. To see this let $k_n=[2n\pi+\frac{\pi}{2}]$ and $l_n=[2n\pi+\frac{3\pi}{2}]$ where $[x]$ denotes the integer part. Then, $a_{k_n}> 0$ while $a_{l_n}<0$. Two subsequences have different limits and so, $(a_n)$ diverges
