Does $a_n \to 0 \iff |a_n|^2 \to 0 $ hold for any sequence? I have tried using the standard definition of convergence
$$
a_n \to 0 \iff
\forall\varepsilon > 0 ~\exists N_{\varepsilon} \in \mathbb{N}:
|a_n| < \varepsilon ~\forall n \geq N_{\varepsilon}
$$
in proving "$\Longrightarrow$". 
Can I argue that for all $n \geq N_{\varepsilon}$ the following holds $|a_n| < \varepsilon \implies |a_n|^2 < \varepsilon$, so $|a_n|^2$ converges?
Is "$\Longleftarrow$" similarly simple?
I've read kobe's proof here on why $a_n \to a \implies a_n^2 \to a^2$ holds, but can't figure out how to apply a similar argument to $\Longleftarrow$.
 A: Yes, and it's all mainly precalculus and basic calculus.
For $\rightarrow$,
$$\lim (|a_n|)^2 = (\lim|a_n|)^2 = (|\lim a_n|)^2 = (|0|)^2 = 0^2 = 0$$
Here, I make use of the fact that $x^2$ and $|x|$ are continuous on $\mathbb R$.
For $\leftarrow$,
$$\lim(|a_n|)^2 = 0 \to (\lim|a_n|)^2 = 0 \to \lim|a_n| = 0 \to |\lim a_n| = 0 \to \lim a_n = 0$$
Here, I make use of the same fact for proving $\rightarrow$ in addition to that $x^2=0 \to x=0$ and that $|x|=0 \to x=0$.

Actually for $a_n \to a \implies a_n^2 \to a^2$, we can just do
$$\lim (a_n^2) = (\lim a_n)^2$$
You can do $\varepsilon$-$N$, if you want, but you're reinventing the wheel a lot. Are you specifically required to do a $\varepsilon$-$N$ proof or forbidden from using that $x^2$ and $|x|$ are continuous on $\mathbb R$?
A: Note that
$$a_n \to 0 \implies |a_n| \to 0$$
and since eventually $|a_n|\le 1$, by squeeze theorem we have that
$$0\le  |a_n|^2 \le|a_n| \to 0$$
Note also that
$$|a_n|^2\to 0 \implies |a_n| \to 0$$
and
$$-|a_n|\le a_n \le |a_n|$$
A: If $|a_n|^2 \to 0$, then $|a_n| = \sqrt{|a_n|^2} \to \sqrt{0} = 0$, by continuity of the root function.
Clearly, then also $a_n \to 0$
