# Proof for convergence of absolute value.

• Oct 20th 2008, 04:05 PM
lllll
Proof for convergence of absolute value.
Prove: $\displaystyle |a_n| \rightarrow 0 \Rightarrow a_n \leftarrow 0$

So what I have is:

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

$\displaystyle 0<|a_n|< \epsilon$

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

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

Originally Posted by lllll
Prove: $\displaystyle |a_n| \rightarrow 0 \Rightarrow a_n \rightarrow 0$

So what I have is:

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

$\displaystyle 0<|a_n|< \epsilon$

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

I'm not quit sure that this is right

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

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

$\displaystyle ||a_n||< \epsilon$

$\displaystyle |a_n|< \epsilon$

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

and that's the exact definition of $\displaystyle a_n \rightarrow 0$ :p
• Oct 21st 2008, 05:49 PM
poutsos.B
The sequence of formulas is as follows:

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

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

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

The above equivalences hold because:

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

$\displaystyle ||a_n|-0|$ =$\displaystyle |a_n-0|$,