Let {an} be a monotone sequence.  Show that {an} converges if and only
if the sequence {an^2) converges.  Is the result true if we onit that
{an} is monotone

Hello fmunshi!
We have to prove that, given {an} is a monotone sequence, then:

1) {an} converges implies {an^2} converges, and
2) {an^2} converges implies {an} converges.

If we can prove these two claims, we will have that {an} converges if
and only
if the sequence {an^2) converges.


Proof of (1)
{an} converges ==> {an} is a Cauchy sequence ==> {an*an} is a Cauchy
sequence ==> {an^2} is a Cauchy sequence ==> {an^2 converges}

Proof that a convergent sequence is Cauchy and that a Real Cauchy
sequence is convergent can be found at

Cauchy sequences
http://www.gap-system.org/~john/analysis/Lectures/L10.html

I've noticed that you've asked in a previous question to prove that
the product of two Cauchy sequences is a Cauchy sequence. I've found
the same resource as websearcher-ga to prove this (that is, to prove
that {an} is a Cauchy sequence ==> {an*an} is a Cauchy sequence)

Sequences and Limits
http://www.math.colostate.edu/~estep/education/517/notesonreals.pdf


Proof of (2)
Since {an^2} is convergent, then it is bounded (above and below).
Proof of this can be found at:

Some properties of convergent sequences
http://www.gap-system.org/~john/analysis/Lectures/L7.html#bdd

Boundedness implies that there exists numbers L and U such that an^2 <
U and an^2 > L for all n. Now:

an^2 < U
|an| < U^(1/2)
-U^(1/2) < an < U^(1/2) for all n

This means that {an} is bounded too. But, we can now use the fact that
given that {an} is bounded and monotone, then it is convergent. Proof
of this is at:

Monotonic sequences
http://www.gap-system.org/~john/analysis/Lectures/L8.html

You can also see a graphical intuition to this proposition at

Monotonce convergence
http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node27.html

So, we have now also proved (2)

Finally, does this result still hold if we omit that {an} is monotone?
The answer is NO. Consider the following {an} sequence, which is not
convergent.
{-2, 2, -2, 2, -2, 2, ...}. 

It's clear that {an^2} is convergent (since this sequence would be
{4,4,4,4,4,4,...} ). This contradicts the hypothesis that {an} is
convergent if {an^2} is convergent.


Best wishes!
elmarto

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Clarification of Answer by elmarto-ga on 10 Jun 2003 15:49 PDT

I forgot to include my search strategy. I searched Google for:
"monotone sequence"
monotone convergent sequence
Cauchy sequence


Best luck!
elmarto