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Let $\{a_{n}\}$ be a sequence of real numbers, where $0

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Yes: since the sequence converges to $0$, for every $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that for all $n\geq N$, we have $0\lt a_n\lt\epsilon$.

So define the sequence recursively: take $a_1$. Then let $\epsilon = \frac{a_1}{2}$; we know there is an $n_1\gt 1$ such that $a_{n_1}\lt \frac{a_1}{2}$. Now take $\epsilon = \frac{a_{n_1}}{2}$; we know there exists $n_2\gt n_1$ such that $a_{n_2}\lt \frac{a_{n_1}}{2}$.

Lather, rinse, and repeat.

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    According to your proof this result is true for any sequence $\{a_{n}\}\subset (0,\infty)$ which converges to 0, not just because $a_{n}\in(0,1)$, I'm right!?2012-06-11
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    Tiny variable name fix (it is not letting me edit it because the edit is not long enough). Where you say "... we know there is an $n_2 > 1$ such ...", that supposed to be $n_1 > 1$ rather than $n_2 > 1.$2012-06-11
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    @Stanley: Yes, you are right.2012-06-11
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Yes. Put $n_1 = 1$. Suppose $n_1 n_k$ so that $a_{n_{k+1}} < \min\{a_{n_j}, 1\le j \le k\}$. Such a sequence will fulfill your specification.

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    Can be simplified slightly, since $\min \{a_{n_j}, 1 \le j \le k\} = a_{n_k}$.2012-06-11