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Alexmahone
Active member
- Jan 26, 2012
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Prove that $a_n\to L\implies|a_n|\to|L|$. (Make cases and use Theorem 5.3B)
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?
My attempt: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 $n\gg 1$
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?