About divergence of sequence Consider the sequence ${x_m}=sin(m)$ .I want to prove this sequence is divergent. So i consider the subsequence $x_{m_k}$ where the sequence  of natural numbers ${m_k}$  is defined as $ m_1=90,   m _{k+1}=360+m_k$ i.e we get subsequence ${(1,1,1,1,.......)}$ which converge to 1 Now I define a new subsequence $x_{m180}=(0,0,0,0....)$ Which converge to  0 so the given sequence become divergent where $m€N$. Is my proof is correct .  if it is correct then tell any improvement in it....
 A: As pointed out you've used non-natural arguments which invalidates the proof. However the basic idea can be used.
The idea you're using is that you have two subsequences that converge to different limits which is enough to prove that the whole sequence doesn't converge. The basic reason for this is that the sequence takes values in two distincts regions. In your example you get that there are values in both $(1-\epsilon,1+\epsilon)$ and $(-\epsilon, +\epsilon)$. The important thing here is not that these intervals are very small, but only that they are disjoint intervals.
To fix the proof you instead construct two subsequences that only has positive values with a marigin and one that only has negative with a marigin. That is that there is a $L>0$ so that one subsequence is always $>L$ and the other is $<-L$.
To see that this sequence exist you consider the range of $\sin(x)$ in the interval $[\pi/2-1/2, \pi/2+1/2]$ in which you certainly can find an integer. There $\sin(x)\ge\sin(\pi/2-1/2) > 0$. So you can find a subsequence that's always at least $\sin(\pi/2-1/2)$ and one that is at most $-\sin(\pi/2-1/2)$. 

A more advanced approach is that you can always find arbitrary large integers that is arbitrarily close to $2n\pi + \xi$ for any $\xi$. This means that while you can't find a subsequence that is always $1$, you can find one that at least approaches $1$. The same is true for $0$.
