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Let $a_{n}$, $b_{n}$, be two sequences of positive real numbers such that $b_{n}< a_{n}$ for all $n\geq 1$, and $a_{n}\leq A$, for some $A>0$ and all $n$. I'm trying to define a new sequence $c_{n}$ in terms of $a_{n}$, $b_{n}$, or both such that

(1) $b_{n}c_{n}\to 0$, and

(2) $a_{n}c_{n}$ does not converges to $0$,

(3) $a_{n}c_{n}$ is bounded above (by something doesn't converge to 0)

I tried $c_{n}=\frac{(a_{n}+1)(b_{n}+2)}{(a_{n}+2)(b_{n}+1)}$, in this case (1) and (3) are true but (2) is not! Anyone has an idea?

Edit:I have a typo! I should write $\{b_{n}\} \subset \{a_{n}\}$, not $b_{n}< a_{n}$ Sorry!!

Also, $a_{n}\to 0$.

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(1) and (2) are a lot to ask if you allow $b_n=a_n$. – Phira Jun 10 '12 at 0:34
We can ignore this possibility. – Nick Jun 10 '12 at 0:36
Wfhat can one ignore? $a_n=2b_n$? What else? – André Nicolas Jun 10 '12 at 0:41
No, just as I said. – Nick Jun 10 '12 at 0:43
You’re going to have a hard time if $\liminf_n\frac{b_n}{a_n}>0$. – Brian M. Scott Jun 10 '12 at 0:43

It is not hard to see that $c_n=\frac1{a_n}$ is the best you can do to have conditions (2) and (3) and the best possible chance for (1).

So, either $b_n/a_n$ converges to 0 and you have your sequence, or it doesn't and your sequence does not exist.

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We don't know if $\frac{b_{n}}{a_{n}}$ converges to 0. – Nick Jun 10 '12 at 0:48
@Nick, that is precisely the point. You cannot expect to construct this $c_n$ unless you know that $b_n/a_n \to 0$. – Erick Wong Jun 10 '12 at 5:32

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