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I am trying to prove: $\inf\{|s_n| : n \in \mathbb{N}\} > 0$. Where $s_n \neq 0 \land \lim s_n \neq 0 \land s = \lim s_n$. My understanding of infimum is that it is the greatest lower bound for a set. My intuition is as the following:

  1. $s_n$ converges implies that that the infimum and supremum exist since there are finitely many points outside the convergence interval $L-\epsilon < s_n < L +\epsilon$.
  2. Since $s_n \neq 0$ Then the absolute value of the infimum is greater than $0$.

Correct me if I am wrong. I have no idea how to formalize this, other than:

Let $\epsilon = \dfrac{s_n}{2}$ then since $s_n \to s$ we can write:

$|s_n -s| < \dfrac{|s|}{2}$ By the definition of the limit

Thus for $n > N \implies |s_n| > \dfrac{s}{2}$ We are in our convergence interval

Therefore there are finitely many points less than $\dfrac{|s|}{2}$ and $\inf\{s_N\} \neq 0$. So, the infimum of $|s_n|$ must be greater than $0$

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    Do you know that $( s_n )$ converges?2012-10-05
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    The assumptions are not clearly stated. It looks as if one of them is that $(s_n)$ converges. And you want to pick a *fixed* $\epsilon$, probably you intend $|s|/2$, not the unreasonable $s_n/2$.2012-10-05
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    @CodeKingPlusPlus: Please check my solution for your question.2012-10-05
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    @ArthurFischer Yes, $s_n$ does converge. I just fixed it in the question.2012-10-07

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