Let $(s_n)$ be a convergent sequence of real numbers such that $s_n \neq 0$ for all $n \in \mathbb{N}$ and $\lim_{n \to \infty}s_n=s\neq 0$.

Question

Prove that $\sup \{\frac{1}{|s_n|} : n \in \mathbb{N}\}>0$

Any help on getting this proof started would be appreciated. I know it must be related to proving that $\inf \{|s_n|:n \in \mathbb{N}\}>0$ but the $\frac{1}{|s_n|}$ is what is throwing me off.

Answer 1

Let $l := \sup_{n \geq 1}|s_{n}|^{-1}$ and suppose $l \leq 0$. Then for every $\varepsilon > 0$ there is some $n \geq 1$ such that $$ l - \varepsilon < |s_{n}|^{-1} \leq l \leq 0. $$ But, since $|s_{n}|^{-1} > 0$ for all $n \geq 1$, $\to \gets$.

Answer 2

Note that a convergent sequence is bounded.

Let $K>0$ denote the bound of $s_n$ so that ${1 \over |s_n|} \ge {1\over K} > 0$ for all $n \in \mathbb N$.

Since this inequality holds for all $n$ it follows that $$\sup_{n \in \mathbb N}{1 \over |s_n|} \ge {1\over K} > 0$$