# Proof for convergence of absolute value.

• Oct 20th 2008, 04:05 PM
lllll
Proof for convergence of absolute value.
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

I'm not quit sure that this is right
• Oct 21st 2008, 11:30 AM
Moo
Hello,
Quote:

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

I'm not quit sure that this is right

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

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 $a_n \rightarrow 0$ :p
• Oct 21st 2008, 05:49 PM
poutsos.B
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:

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