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Prove that |a_n| tends to |L|

Alexmahone

Active member
Jan 26, 2012
268
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
Jan 26, 2012
236
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
Jan 26, 2012
268
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
Jan 26, 2012
890
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
Jan 26, 2012
268

CaptainBlack

Well-known member
Jan 26, 2012
890
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 \(|a_n|\to 0=|L|\)