2
$\begingroup$

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, $a_{n}\to 0$.

  • 0
    (1) and (2) are a lot to ask if you allow $b_n=a_n$.2012-06-10
  • 0
    We can ignore this possibility.2012-06-10
  • 0
    Wfhat can one ignore? $a_n=2b_n$? What else?2012-06-10
  • 0
    No, just as I said.2012-06-10
  • 0
    You’re going to have a hard time if $\liminf_n\frac{b_n}{a_n}>0$.2012-06-10
  • 0
    Oh, I have a typo! I should write $\{b_{n}\} \subset \{a_{n}\}$, not $b_{n}< a_{n}$.2012-06-10
  • 0
    Andre's question is still valid.2012-06-10

1 Answers 1