# Convergent sequence

1. Sep 30, 2009

### Daveyboy

For a sequence in the reals

{an} converges to a, show {|an|} converges to |a|.

For any e>0 the exists an N s.t. for any n>N |an-a|<e

I want to use this inequality, but there is something funny going on. I do not know how to justify it.

|an-a|$$\leq$$||an|-|a||

2. Sep 30, 2009

### HallsofIvy

Staff Emeritus
Look at three separate cases.

1) a> 0. Can you show that, for some N, for all n> N $a_n> 0$? (Take $\epsilon= a/2$.)

2) a< 0. Can you show that, for some N, for all n> N $a_n< 0$?

3) a= 0. Here, [itex]||a_n|- a|= ||a_n||= |a_n|.

3. Sep 30, 2009

### Daveyboy

Okay I see how to break it down case wise and find N accordingly. That will work nicely.

However, I was hoping to use the reverse triangle inequality but I run into the double abs. value. It just doesn't look right to say that
for any e>0 there exists and N s.t. for any n >N

|an - a| < e
and
|an - a| $$\geq$$ |an| - |a|
implies

e>||an| - |a||

but if I showed this wouldn't it be true?