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?


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.

  • $\begingroup$ I like the proof because it is so short but why can we write $b_n=(b_n/a_n)a_n$? Or more generaly if we know that a sequence is a (finite) composition of convergent sequences why this sequnce must be also convergent? $\endgroup$ – New2Math Mar 19 at 14:12
  • $\begingroup$ This notation may make it clearer: $b_n=\dfrac{b_n}{a_n}a_n.$ I am assuming you know the product rule for convergent sequences. We know $b_n/a_n\to 1$ and $a_n\to a.$ Therefore $b_n=\dfrac{b_n}{a_n}a_n \to 1\cdot a.$ $\endgroup$ – zhw. Mar 19 at 15:52
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    $\begingroup$ Can I ask the downvoter to explain the downvote? $\endgroup$ – zhw. Mar 19 at 15:53

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 $1$

  • $\begingroup$ What do you mean with if we go far enough out that $b_n\in[1/2,2]$? $\endgroup$ – New2Math Mar 16 at 14:12
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    $\begingroup$ It means that there is some $N$ such that $n \gt N \implies b_n \in [\frac b2,2b]$. Note I had forgotten the factor $b$ in it. I am using the fact that $b_n \to b$ to make sure it is close enough to be within a factor $2$ so I can use the value $\frac \epsilon 2$ $\endgroup$ – Ross Millikan Mar 16 at 14:43
  • $\begingroup$ @New2Math: I can't tell from the comments what you do and don't understand. The fact that comments cannot be edited after 5 minutes makes it hard. It would be best if you edit an addendum to the question saying Added: I understand this and don't understand that. Alternately, write one new comprehensive comment and delete the earlier ones. Thanks $\endgroup$ – Ross Millikan Mar 16 at 20:57
  • $\begingroup$ What I understood was that $|a_n-b|>\epsilon$ for infinitely many $n$. I also understood that since $b_n\rightarrow b$ and if $b\neq 0$ we can say for $n>N\quad |b_n-b|<\frac{1}{2}|b|\Rightarrow |b_n|-|b|<\frac{|b|}{2}\Rightarrow |b_n|<|b|\frac{3}{2}<2|b|$ and $|b|-|b_n|<\frac{1}{2}|b|\Rightarrow |b_n|>\frac{1}{2}|b|$. Therefore $|b_n|\in[\frac{|b|}{2},2|b|]$. So I understood the preperations but I don't know how I can prove now with those preperations that $|\frac{a_n}{b_n}-1|>\frac{\epsilon}{2}$. $\endgroup$ – New2Math Mar 17 at 11:44
  • $\begingroup$ I don't understand how to handle the case if $b=0$ and also the reverse case when $a_n$ is convergent and $b_n$ is divergent. Please elaborate I would give you $50$ rep $\endgroup$ – New2Math Mar 17 at 11:51

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).

  • $\begingroup$ I love it we have proved that if they are asymptoe they necessarily both converge to the same value. $\endgroup$ – New2Math Mar 19 at 13:51
  • $\begingroup$ It does include this case. I’ll update my answer soon to show you how $\endgroup$ – csprun Mar 19 at 14:00
  • $\begingroup$ What I thought is that if $a_n$ and $b_n$ are Asymptote then they are necessarily either convergent to the same value or both divergent. Since this is a necessary condition only it is not sufficient to have to have two sequences which converge to the same value. The Task was to prove the implication not the equivalence. So it is all good. $\endgroup$ – New2Math Mar 19 at 14:04
  • $\begingroup$ My argument works as long as $b_n \ne 0$ for infinitely many $n$ (just ignore the finitely many $n$ where $b_n = 0$). And that's a hidden assumption when you say $a_n/b_n \to 1$, so I'm not assuming anything extra. $\endgroup$ – csprun Mar 19 at 14:21

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