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I have to prove the following:

If $(c_n)_{n\in \mathbb{N}} $ is any real sequence. I should show that there exist two bounded sequences $(a_n)_{n\in \mathbb{N}}, (b_n)_{n\in \mathbb{N}}$ such that $c_n = \frac{a_n}{b_n}$ for all $n \in \mathbb{N} $.

So my guess was that:

if $c_n$ is bounded I can say $a_n=c_n, b_n=1$

if $c_n$ is not bounded $a_n=1, b_n=\frac{1}{c_n}$

So my questions are:

  • Can or should I proof first that if $a_n,b_n$ is bounded that $c_n$ is also bounded?
  • Because if $c_n$ is a $0$ sequence I have a division by $0$

Thank you.

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  • $\begingroup$ By "limited" do you mean "bounded" or "such that $\lim_{n\to\infty} a_n$ exists"? $\endgroup$ – user228113 Mar 12 '17 at 15:14
  • $\begingroup$ I mean that they have a supremum (infimum). $\endgroup$ – flow Mar 12 '17 at 15:16
  • $\begingroup$ Ok. It is "bounded". $\endgroup$ – user228113 Mar 12 '17 at 15:18
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  • It is not true that $(a_n)_{n\in\Bbb N},(b_n)_{n\in\Bbb N}\text{ limited}\implies\left(\frac{a_n}{b_n}\right)_{n\in\Bbb N}\text{ limited}$. For instance, consider $a_n=1,\ b_n=\frac1n$

  • In fact, you grasped one part of the idea, but not the other one, which, allows you to use it effectively. To complete it, set a given number $u>0$. Observe that, for the values of $n$ such that $\lvert c_n\rvert>u$, you have $-\frac1u<\frac1{c_n}<\frac1u$. So that subsequence is bounded. On the other hand, when $\lvert c_n\rvert\le u$... Well that's bounded.

So, what's an educated guess of a piece-wise definition of $a_n$ and $b_n$ ?

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  • $\begingroup$ For the definition of $a_n, b_n$ bounded I would use: $a_n \le S_a$ and $b_n \le S_b$ => $a_n + b_n \le S_a + S_b$ $\endgroup$ – flow Mar 12 '17 at 15:29
  • $\begingroup$ And the definition for the sequence $|a_n - a| < \epsilon_a, |b_n - b| < \epsilon_b$ $\endgroup$ – flow Mar 12 '17 at 15:41
  • $\begingroup$ @flow Are those strings of symbols supposed to be related to the problem at hand? $\endgroup$ – user228113 Mar 12 '17 at 17:32
  • $\begingroup$ I don't understand, thank you anyway. $\endgroup$ – flow Mar 13 '17 at 9:38

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