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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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    @ArthurFischer Yes, $s_n$ does converge. I just fixed it in the question.2012-10-07

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In the following we assume that $s_n\ne 0$ for all $n\in \mathbb{N}$. Let $\displaystyle s=\lim_{n\rightarrow\infty}s_n\ne 0$. Then $ \lim_{n \rightarrow \infty}|s_n|=|s|>0. $ Let $\displaystyle\varepsilon =\frac{|s|}{2}$. Then there exists $N\in\mathbb{N}$ such that $ ||s_n|-|s||<\frac{|s|}{2} \quad \forall n\geq N. $ Hence $ |s_n|>\frac{|s|}{2} \quad \forall n\geq N. $ Let $\bar{s}=\min\left\{|s_1, |s_2|, \ldots, \frac{|s|}{2}\right\}>0,$ then $|s_n|\geq \bar{s}$ for all $n\in \mathbb{N}$. So $ \inf\{|s_n|:\in\mathbb{N}\}\geq \bar{s}>0. $

Note. If there exists $n_0\in \mathbb{N}$ such that $s_{n_0}=0$ then $ 0\leq \inf\{|s_n|: n\in \mathbb{N}\}\leq |s_{n_0}|= 0, $ which implies that $ \inf\{|s_n|: n\in \mathbb{N}\}=0. $