Proof for convergence of absolute value.

**lllll** · October 20th 2008, 04:05 PM

Prove: $|a_n| \rightarrow 0 \Rightarrow a_n \leftarrow 0$

So what I have is:

$0<||a_n|-0|< \epsilon$

$0<|a_n|< \epsilon$

now $a_n \leq |a_n| \therefore 0<a_n \leq |a_n|< \epsilon \therefore 0<a_n < \epsilon$

I'm not quit sure that this is right

**Moo** · October 21st 2008, 11:30 AM

Hello,

Originally Posted by lllll

Prove: $|a_n| \rightarrow 0 \Rightarrow a_n \rightarrow 0$

So what I have is:

$0<||a_n|-0|< \epsilon$

$0<|a_n|< \epsilon$

now $a_n \leq |a_n| \therefore 0<a_n \leq |a_n|< \epsilon \therefore 0<a_n < \epsilon$

I'm not quit sure that this is right

Hmm you have to separate the cases when $a_n<0$ and $a_n>0$

More simply :
$|a_n| \to 0 \Leftrightarrow \forall \epsilon >0, ~ \exists N \in \mathbb{N}, ~ \forall n>N ~:$
$||a_n|-0|< \epsilon$

$||a_n||< \epsilon$

$|a_n|< \epsilon$

$|a_n-0|< \epsilon$

and that's the exact definition of $a_n \rightarrow 0$

**poutsos.B** · October 21st 2008, 05:49 PM

The sequence of formulas is as follows:

Lim $|a_n|$ =0 <====> $\forall\epsilon$ [ε>0------> $\exists k$ ( kεN & $\forall n$ ( n $\geq$ k--------> $||a_n|-0|< \epsilon$ ))] <=====>

$\forall\epsilon$ [ε>0------> $\exists k$ ( kεN & $\forall n$ ( n $\geq$ k--------> $|a_n-0|< \epsilon$ ))] <======> Lim $(a_n)$ =0

As n-----> OO ( infinity).

The above equivalences hold because:

$||a_n|-0|< \epsilon$ <=====> $|a_n-0|< \epsilon$ , since:

$||a_n|-0|$ = $|a_n-0|$ ,

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