A divergent sequence and a convergent sequence cannot be asymptotic equal Two sequences $(a_n),(b_n)$ are called asymptotic equal. If the sequence $(a_n/b_n)$ converges to 1
$$\frac{a_n}{b_n}\rightarrow1,n\rightarrow\infty\quad a_n\cong b_n\,n\rightarrow\infty$$
According to the rule (if $a_n\rightarrow a,b_n\rightarrow b\neq 0\quad \frac{a_n}{b_n}\rightarrow\frac{a}{b}$) asymptotic equal sequences are either convergent (in the same manner) at the same time or divergent (in the same manner) at the same time. 
I don't know how the rule that I have stated implies that a divergent sequence and a convergent sequence cannot be asymptotic equal.
Can somebody explain me this?  

Let $a_n$ be divergent and $b_n$ be convergent.  If it is the reverse you can invert the factions.  Let $b_n \to b.$  Assume $b \gt 0$, otherwise negate the terms to make it so.  Saying $a_n$ is divergent means that for any proposed limit $a$, there is some $\epsilon \gt 0$ such that there are infinitely many $n$ such that $|a_n-a|\gt \epsilon$.  In particular, there are infinitely many $n$ such that $|a_n - b| \gt \epsilon$.  We assume $\epsilon \ll 1$, because if it isn't we can shrink it to be so.  If we go far enough out that $b_n \in [b-\epsilon^2,b+\epsilon^2]$ then we have $|\frac {a_n}{b_n} -1| \gt \frac \epsilon {b}$ for those $n$ where $|a_n - b| \gt \epsilon$, so $\frac {a_n}{b_n}$ is not convergent to $1$ 
Added:  if it is $b_n$ that diverges, the argument works the same.  Let $a_n \to a$ and eventually we have $\frac a2 \lt a_n \lt 2a$.  There is some epsilon such that infinitely many $b_n$ such that $|b_n-a| \gt \epsilon$  Then $|\frac {a_n}{b_n}-1| \gt |\frac a{a+\epsilon}-1|\sim \frac \epsilon a$ 
If the limit of the convergent sequence is $0$ the argument is even stronger because of the $a$ or $b$ in the denominator.  You can just say the error is greater than $\epsilon$ whenever the limit is less than $1$
