Does Convergence of a Sequence Imply Convergence of Its Absolute Values?

[alexmahone]

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]

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]

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]

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]

Case 3: \(L=0\) left to the reader

I don't see how this case can be tackled like the first 2 cases.

 

[CaptainBlack]

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 \(|a_n|\to 0=|L|\)