Does $a_n \to 0 \iff |a_n|^2 \to 0 $ hold for any sequence?

Question

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 $\Longleftarrow$.

Answer 1

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 $|x|=0 \to x=0$.

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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 $\mathbb R$?

Answer 2

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

$$-|a_n|\le a_n \le |a_n|$$

Answer 3

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 $a_n \to 0$