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?
Best Answer
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.