# The sequence $(|a_n|)$

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Theorem
If $a_n\to a$, then $|a_n|\to |a|$.

Proof
Given $\varepsilon >0$, there exists $N$ such that $|a_n-a|\lt \varepsilon$ for all $n>N$. But by the reverse triangle inequality
$\left| |a_n|-|a| \right| \leq |a_n-a| \lt \varepsilon {\text{ for all }} n>N.$
Hence, $|a_n|\to |a|$.

This allows us to prove a simple result which turns out to be very useful.
Corollary
If $a_n\to a$ and $a_n\geq 0$ for all $n\in \N$, then $a\geq 0$.

Proof
From the theorem we have that $a_n=|a_n|\to |a|$ and since limits are unique $a=|a|$, i.e., $a\geq 0$.

We give an alternative proof for the corollary which rather than use the theorem uses ‘first principles’. I’ve included this as I have seen many first principles proofs in books and online that make it much more complicated than it needs to be.

Alternative Proof
Assume for a contradiction that $a\lt 0$. Then as $-a>0$, there exists $N$ such that
$|a_n-a|\lt -a {\text{ for all }} n>N.$
Then, as $a_n-a>0$ (because $a_n\geq0$ and $a\lt 0$), we have for $n>N$,
$a_n-a=|a_n-a|\lt -a ,$
i.e., $a_n\lt 0$, which is a contradiction.

Hence, $a\geq 0$.

How to think like a mathematician
As usual we ask `What about the converse?’. Well, the converse to the theorem is not true. We can have $|a_n|\to |a|$ but $a_n\not\to a$. For example, $a_n=(-1)^n$.

The converse is true in a very special case (and we will make use of this a number of times):

Theorem
If $|a_n|\to 0$, then $a_n\to 0$.

Proof
Given $\varepsilon >0$, there exists an integer $N$ such that $n>N$ implies $\left| | a_n| – 0 \right|<\varepsilon$ for all $n>N$ since $|a_n|\to 0$.

But, $|a_n-0|=|a_n|=\left| | a_n| – 0 \right|\lt \varepsilon$ so $a_n\to 0$ also.