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We know the sequence space $c_{o}$, which is the space of all sequences converging to zero, and it is a Banach space. Consider a subset $S$ of $c_{o}$ of (some) such sequences which are decreasing to 0, i.e., $S=\left\{\{s_{n}\}: \{s_n\}\; \text{is decreasing and}\; s_{n}\to 0\right\}$.

Can we find a sequence $\{s^'_{n}\}\in S$ such that $$\lim_{n\to\infty}\frac{s^'_{n}}{s_{n}}=0$$ for all sequences $\{s_{n}\}\in S$? (other than $s^'_{n}$)

(I've posted a similar question but in another form, but I re-formulate it in a way to be easy to understand, and contains the idea I want. So, since I didn't get answers for the old question it can be deleted)

My previuos guess was to take $s^'_{n}$ to be the product of all sequences in $S$, but I'm not sure if this is correct!

Any help is appreciated!

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2 Answers

If I understand the question correctly, then no, such a sequence $s'$ cannot exist, as if there were such a sequence, then surely we could take a subsequence $s''$ of $s'$ which decreases sufficiently more rapidly to zero, and then $\lim_{n\to\infty}\frac{s'_{n}}{s''_{n}}=0$.

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Is it possible to multiply the ratio $\frac{s'}{s}$ by a fixed sequence $x_{n}$ converging to zero, not necessary from $S$, to make the problem true? –  Paul Jun 12 '12 at 11:07
    
Unfortunately, no. That would really be equivalent to replacing the sequence $\{s'_n\}$ by a new sequence $\{x_ns'_n\}$, and we have already decided that such a sequence cannot have the property you want. –  Old John Jun 12 '12 at 11:13
    
So is there any case where such sequence could exist? For example, any conditions on $S$ ? –  Paul Jun 12 '12 at 11:16
    
I can't think of anyway you could get such a sequence. You are really asking for a sequence which converges to zero more rapidly than any other sequence, and we have proved that such a sequence is impossible. –  Old John Jun 12 '12 at 11:18
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No, you can’t: for any $s\in S$, the sequence $t=\left\langle\frac{s_n}{2^n}:n\in\Bbb N\right\rangle$ has the property that $$\lim_{n\to\infty}\frac{t_n}{s_n}=0\;.$$ Informally, there’s always a strictly ‘smaller’ sequence.

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But I think at some point in the process of taking subsequences all the subsequences will behaves in a same rate, is it true? –  Paul Jun 12 '12 at 11:15
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@Paul: I’m not sure what you mean: the question isn’t about subsequences. But even if it were, the answer would be no: if a sequence decreases with limit $0$, in any subsequence you can find a sub-subsequence that converges faster. –  Brian M. Scott Jun 12 '12 at 11:17
    
Is it true the other way; given a sequence $s$, there is a sequence $s'$ such that $s\subset s'$ and $\lim s/s'=0$? –  Paul Jun 12 '12 at 11:29
    
@Paul: Yes: you just have to add a lot of terms, so that $s'_n$ decreases more slowly than $s_n$. –  Brian M. Scott Jun 12 '12 at 11:33
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