Math Is Fun Forum

cxc001

Member

Registered: 2010-04-09

Posts: 17

Convergence of Sequence "Does {An^2} converges => {An} converges?

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 n^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?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Convergence of Sequence "Does {An^2} converges => {An} converges?

I kinda think it is false, but couldn’t think of any counterexample to directly proof it.

cxc001

Member

Registered: 2010-04-09

Posts: 17

Re: Convergence of Sequence "Does {An^2} converges => {An} converges?

Got it, thanks!