Prove $S_n \rightarrow 0 \iff |S_n| \rightarrow 0$

Question

Just a quick and supposedly simple question which I don't understand how to do.

How would you prove $$S_n\to 0 \iff|S_n|\to 0$$

Answer 1

Just playing with definition $$ \begin{align} \lim\limits_{n\to\infty}S_n=0&\Longleftrightarrow (\forall\varepsilon>0\quad\exists N\in\mathbb{N}\quad\forall n>N\implies|S_n-0|\leq\varepsilon)\\ &\Longleftrightarrow (\forall\varepsilon>0\quad\exists N\in\mathbb{N}\quad\forall n>N\implies||S_n|-0|\leq\varepsilon) \Longleftrightarrow\lim\limits_{n\to\infty}|S_n|=0\end{align} $$

Answer 2

One way is obvious, for the other way use that $|x|=0\iff x=0$ and that $| \cdot | $ is continuous.

Or more elementary use $$-|S_n|\leq S_n \leq |S_n|$$

Answer 3

$-|S_n|\le S_n\le|S_n|$

So if $|S_n|\to 0 $ as $n\to \infty $ then $S_n\to 0$ also.

Answer 4

If $a_n\to 0$ so $$\forall~\epsilon>0,~\exists N\in\Bbb N,~~n>N \Longrightarrow |a_n|<\epsilon$$ This equals to $$\forall~\epsilon>0,~\exists N\in\Bbb N,~~n>N \Longrightarrow \big||a_n||<\epsilon$$ This equals to $|a_n|\to 0$