Convergence of Sequence Does {An^2} converges => {An} converges? How to prove it?

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Convergence of Sequence "Does {An^2} converges => {An} converges? How to prove it?"

Does sequence {An^2} converges implies to sequence {An} converges? True or False. How to prove it?

I kinda think it is false, but couldn’t think of any counterexample to directly proof it. So I try to use the 1) definition of convergence and 2) the Comparison Lemma to prove it, but kinda stucked.

Proof1: (Use definition of convergence)
Let sequence An^2 converges to a^2
Then according to the definition of convergence
For every E>0, Find N such that
|An^2-a^2|<E for all n>N
|(An-a)(An+a)|=|An-a||An+a|<E
|An-a|<E/|An+a|)

How can I go from here?

So if I can let N=E/|An+a|+1, then An converges to a. But I can’t define N that has a sequence in it, can I?

Comparison Lemma states “Let sequence {An} converges to a, and let {Bn} be a sequence such that |Bn-b|<= C|An-a| for some C>0, then Bn converges to b”

Proof2: (Use Comparison Lemma)
Let sequence An^2 converges to a^2
Let {An} be a sequence such that |An-a|<=C|An^2-a^2| for some C>0
Now I need to find an C>0 so that I can prove that sequence An converges to a

How can I go from here?
 

Answers and Replies

  • #2
mathman
Science Advisor
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Your first proof looks OK unless a=0.
 
  • #3
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1, -1, 1, -1, 1, -1, ...
 
  • #5
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So choose sequence An^2 = [(-1)^n]^2 and the sequence An^2 converges to 1, but An is NOT convergent (divergent) sequence. Super! Thanks!
 

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