# Prove that |a_n| tends to |L|

#### Alexmahone

##### Active member
Prove that $a_n\to L\implies|a_n|\to|L|$. (Make cases and use Theorem 5.3B)

Theorem 5.3B: Assuming $\{a_n\}$ converges,

$\lim\ a_n<M\implies a_n<M$ for $n\gg 1$

$\lim\ a_n>M\implies a_n>M$ for $n\gg 1$
My attempt:

Given $\epsilon>0$,

$L-\epsilon<\lim\ a_n<L+\epsilon$ for all $n$

$L-\epsilon<a_n<L+\epsilon$ for $n\gg 1$ (Using Theorem 5.3B)

$-\epsilon<a_n-L<\epsilon$ for $n\gg 1$

$|a_n-L|<\epsilon$ for $n\gg 1$

Case 1: $|a_n|\ge|L|$ for $n\gg 1$

$|a_n|-|L|\le|a_n-L|<\epsilon$ for $n\gg 1$

So, $|a_n|\to|L|$

Case 2: $|a_n|<|L|$ for $n\gg 1$

$|L|-|a_n|\le|a_n-L|<\epsilon$ for $n\gg 1$

So, $|a_n|\to|L|$

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Could someone please check the above proof for me?

#### ThePerfectHacker

##### Well-known member
Prove that $a_n\to L\implies|a_n|\to|L|$. (Make cases and use Theorem 5.3B)

My attempt:

Given $\epsilon>0$,

$L-\epsilon<\lim\ a_n<L+\epsilon$ for all $n$

$L-\epsilon<a_n<L+\epsilon$ for $n\gg 1$ (Using Theorem 5.3B)

$-\epsilon<a_n-L<\epsilon$ for $n\gg 1$

$|a_n-L|<\epsilon$ for $n\gg 1$
This looks strange for a proof. Basicaly what you proved is that $|a_n - L | < \varepsilon$ for $n$ sufficiently large. But we know that already as that is the definition of convergence! You did not prove anything useful, that is the very definition.

#### Alexmahone

##### Active member
This looks strange for a proof. Basicaly what you proved is that $|a_n - L | < \varepsilon$ for $n$ sufficiently large. But we know that already as that is the definition of convergence! You did not prove anything useful, that is the very definition.
I guess the only thing required is $||x|-|y||\le|x-y|$, which can be proved by squaring both sides.

#### CaptainBlack

##### Well-known member
Prove that $a_n\to L\implies|a_n|\to|L|$. (Make cases and use Theorem 5.3B)

Theorem 5.3B: Assuming $\{a_n\}$ converges,

$\lim\ a_n<M\implies a_n<M$ for $n\gg 1$

$\lim\ a_n>M\implies a_n>M$ for $n\gg 1$
Case 1: $$L>0$$

Then by the theorem the $$a_n$$s are eventually all positive and so from that point on $$|a_n|\to \lim_{n\to \infty} a_n=L=|L|$$

Case 2: $$L<0$$

Then by the theorem the $$a_n$$s are eventually all negative so $$|a_n| \to \lim_{n \to \infty} -a_n=-L=|L|$$

Case 3: $$L=0$$ left to the reader

CB

#### Alexmahone

##### Active member
Case 3: $$L=0$$ left to the reader
I don't see how this case can be tackled like the first 2 cases.

#### CaptainBlack

##### Well-known member
I don't see how this case can be tackled like the first 2 cases.
Case 3 is for $$a_n$$ being a null sequence, which reduces straight off to $$|a_n|\to 0=|L|$$