Convergent sequence

  • Thread starter Daveyboy
  • Start date
  • #1
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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|[tex]\leq[/tex]||an|-|a||
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Look at three separate cases.

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

2) a< 0. Can you show that, for some N, for all n> N [itex]a_n< 0[/itex]?

3) a= 0. Here, [itex]||a_n|- a|= ||a_n||= |a_n|.
 
  • #3
58
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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| [tex]\geq[/tex] |an| - |a|
implies

e>||an| - |a||

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

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