1
$\begingroup$

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!

2 Answers 2

3

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 $\lim_{n\to\infty}\frac{s'_{n}}{s''_{n}}=0$.

  • 0
    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.2012-06-12
2

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.

  • 0
    @Paul: Yes: you just have to add a lot of terms, so that $s'_n$ decreases more slowly than $s_n$.2012-06-12