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?  
 A: 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 $1$
A: In your third line, I can't tell whether you think that $(a_n)$ and $(b_n)$ are both assumed to be convergent in the definition of being asymptotically equal. If so, that might be a large chunk of what's confusing you. But I'll assume you know that's not part of the definition, and I'll go along with an explanation.

Instead of going through a full proof, I think it's better for me to translate the assumptions into something that's much easier to visualize and understand.
Suppose one of $(a_n)$ and $(b_n)$ is convergent. 
First let's show that $(b_n)$ must be bounded. If not, then $(b_n)$ cannot be convergent, so $(a_n)$ must be convergent, and so $(a_n)$ is bounded. But then $(b_n)$ must also be bounded: otherwise, if $b_n$ is unbounded, the ratio $a_n/b_n$ will have infinitely many terms that are too close to $0$ -- for example, infinitely many terms in $(-1/2,1/2)$, so $a_n/b_n$ cannot converge to $1$, a contradiction. So $(b_n)$ is bounded. Let $B \in \mathbb{R}$ be a positive number such that $|b_n|<B$ for all $n\in \mathbb{N}$. 
Now we'll translate the statement about the ratio $a_n/b_n$ into a direct comparison between $a_n$ and $b_n$. Fix $k\in \mathbb{N}$. Let $\epsilon_k = 1/(k B)$. Since $(a_n)$ and $(b_n)$ are asymptotically equal, we have that there exists $n_k \in \mathbb{N}$ so that when $n\ge n_k$, we have 
\begin{equation}\tag{1}\left| \frac{a_n}{b_n}-1 \right|<\epsilon_k.\end{equation}
If $b_n>0$, then $0 < b_n/B < 1$, so (1) can be rewritten as
$$b_n(1-\epsilon_k) < a_n < b_n(1+\epsilon_k),$$
so
$$b_n-1/k < b_n\left(1-\frac{1}{kB}\right) < a_n < b_n\left(1+\frac{1}{kB}\right) < b_n + 1/k.$$
Similarly, if $b_n<0$, then $-1<b_n/B<0$, so (1) can be rewritten as 
$$b_n(1+\epsilon_k) < a_n < b_n(1-\epsilon_k),$$
which similarly implies
$$b_n-1/k < a_n < b_n + 1/k.$$
Just to make it clear, all this only necessarily holds for $n\ge n_k$. So for $n\ge n_k$, we have 
$$b_n-1/k < a_n < b_n + 1/k.$$ This implies the following: for any $k\in \mathbb{N}$, there is some $n_k \in \mathbb{N}$ so that when $n\ge n_k$, you can fit $a_n$ and $b_n$ in an interval of length $2/k$. This makes the problem much easier to visualize for me, and if it works for you too, it should be a lot easier to see why they both must be convergent now (remember, we're assuming one of them already is convergent).
A: Here's my definition of $a_n\sim b_n:$ i) $a_n,b_n$ are nonzero for every $n;$ ii) $a_n/b_n \to 1.$
The requirement i) guarantees both $a_n/b_n$ and $b_n/a_n$ are well defined for all $n.$ Thus if $a_n/b_n \to 1,$ then $b_n/a_n = 1/(a_n/b_n) \to 1$ by the rule you stated. It follows that $a_n\sim b_n$ iff $b_n\sim a_n.$
Thm: If $a_n\sim b_n,$ then $a_n,b_n$ either both converge or both diverge.
Proof: Suppose $a_n$ converges. Note that $b_n = (b_n/a_n)a_n$ for all $n.$ Because $b_n\sim a_n,$ $b_n/a_n \to 1.$ Thus $b_n$ is the product of two convergent sequences. By the product rule for limits, $b_n$ converges to $1\cdot a = a,$ where $a=\lim a_n.$ So $a_n$ convergent implies $b_n$ convergent, and by symmetry, $b_n$ convergent implies $a_n$ convergent.
We've shown that if $a_n\sim b_n,$ then $a_n$ converges iff $b_n$ converges. Could $a_n$ converge while $b_n$ diverges? No. We just showed $a_n$ convergent implies $b_n$ convergent. And of course by symmetry, $b_n$ cannot converge while $a_n$ diverges.
