# Limits of Subsequences

If $s=\{s_n\}$ and $t=\{t_n\}$ are two nonzero decreasing sequences converging to 0, such that $s_n ≤t_n$ for all $n$. Can we find subsequences $s ′$ of $s$ and $t ′$ of $t$ such that $\lim \frac{s'}{t'}=0$ , i.e., $s ′$ decreases more rapidly than $t ′$ ?

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Yes, we can. So we have two positive decreasing sequences with $s_n \leq t_n$, and $s_n \to 0, t_n \to 0$.
Then we can let $t' \equiv t$. As $\{t_n\}$ is positive, $t_1 > 0$. As $s_n \to 0$, there is some $k$ s.t. $s_k < t_1/1$. Similarly, there is some $l > k$ s.t. $s_l < t_2/2$. Continuing in this fashion, we see that we can find a sequence $s'$ so that $s'_n < t_n/n$, so that $\dfrac{s'_n}{t_n} < \dfrac{1}{n}$.
Are you sure about the index of $s'_{n}$ in the last line above? – Paul Jun 12 '12 at 12:49
@Paul: I think you're asking me if I can say $s'_n$. Without writing all the details, I let $s_k =: s'_1$ and $s_l =: s'_2$, and so on. Is that what you meant? – mixedmath Jun 12 '12 at 12:51
With the same idea it actually follows that for any subsequence of $\{t_{n}\}$ we find such a subsequence of $\{s_{n}\}$ that decreases more rapidly. And I don't think we needed the fact that $s_{n}\leq t_{n}$ at all. – T. Eskin Jun 12 '12 at 12:52
So the method of proof works also in the case if $s_{n}=t_{n},\forall n$, right! – Paul Jun 12 '12 at 12:59