1. ## Sequence Convergence

Recall that a sequence ($\displaystyle a_n$) converges to L if and only if

$\displaystyle \forall \epsilon >0 \exists N \in \mathbb{N} \forall n \geq N |a_n - L|< \epsilon$

Prove that if $\displaystyle a_n \rightarrow L$ then $\displaystyle |a_n| \rightarrow |L|$.

Attempt:
Suppose $\displaystyle a_n \rightarrow L$, then $\displaystyle \lim_{x-> \infty} a_n = L$

Then we know $\displaystyle -a_n \leq a_n \leq a_n$

The limit of the two outside terms is L, and hence the limit of $\displaystyle a_n$ is L by the squeezing theorem for sequences.

Conversely we can say: suppose $\displaystyle |a_n| \rightarrow L$ then

$\displaystyle -|a_n| \leq a_n \leq |a_n|$

etc...

But I doubt this is right or even the right approach. Can anyone show me a better and more rigorous proof?

2. Originally Posted by Roam
Prove that if $\displaystyle a_n \rightarrow L$ then $\displaystyle |a_n| \rightarrow |L|$.
It is really simple. This is all you need.
$\displaystyle \left| {\left| {a_n } \right| - \left| L \right|} \right| \leqslant \left| {a_n - L} \right|$

3. That was really simple! Much simpler than what I was thinking:

1) Suppose L is positive. Then you can take $\displaystyle \epsilon$ = L/2 to show that for large enough n, $\displaystyle a_n> 0$ so $\displaystyle |a_n|= a_n$ and the sequence $\displaystyle |a_n|$ converges to L.

2) Suppose that L is negative. Then you can take $\displaystyle \epsilon$= -L/2 to show that for large enough n, $\displaystyle a_n< 0$ so $\displaystyle |a_n= -a_n$ and the sequence $\displaystyle |a_n|$ converges to -L.

3) Suppose that L= 0. Then [tex]||a_n|- |L||= ||a_n||= |a_n| = |a_n- L|.