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Please prove that any sequence converging to a non-zero value is bounded away from 0.

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up vote 2 down vote accepted

This is not true. Suppose $a_1 = 0$ and $a_n = 2$ for all $n > 1$. $\lim_{n \rightarrow \infty} a_n = 2$ but $\{a_n : n \in \mathbb{N}\}$ is not bounded away from $0$.

However if you assume that the sequence $(a_n)_{n \in \mathbb{N}}$ is nonzero, then this is true.

Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers. Suppose that $\lim_{n \rightarrow \infty} a_n = L$ where $L \neq 0$. By the definition of the limit, there exists a natural number $N$ such that for all $n > N$, $|a_n - L| < \frac{|L|}{2}$. Let $\alpha = min \{|a_0|, ..., |a_n|, |\frac{L}{2}|\}$. Then you have that for all $n$, $|a_n| > \alpha$.

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I don't think the first example is correct. I think "bounded away from zero" means that for some $\,\epsilon > 0\,$ there exists an index $\,n\,\,s.t.\,\,|a_m|>\epsilon\,\,\,\forall m\geq n\,$ – DonAntonio Aug 16 '12 at 4:17
Thank you very much sir, that's what I needed. – Zaataro Aug 16 '12 at 4:20
@DonAntonio: $\:$ Your definition would be for "eventually bounded away from zero". $\hspace{1 in}$ – Ricky Demer Aug 16 '12 at 4:52
+1 for a good answer. Wish I could do more for an example that shows the issues well. – Ross Millikan Aug 16 '12 at 5:17
Perhaps it is just a matter of definition @RickyDemer – DonAntonio Aug 16 '12 at 11:10

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