Google Answers Logo
View Question
Q: Sequences of real numbers ( Answered,   0 Comments )
Subject: Sequences of real numbers
Category: Science > Math
Asked by: fmunshi-ga
List Price: $6.00
Posted: 10 Jun 2003 12:13 PDT
Expires: 10 Jul 2003 12:13 PDT
Question ID: 215685
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
Subject: Re: Sequences of real numbers
Answered By: elmarto-ga on 10 Jun 2003 15:48 PDT
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

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

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

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

You can also see a graphical intuition to this proposition at

Monotonce convergence

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
{-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!

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!
There are no comments at this time.

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  

Google Home - Answers FAQ - Terms of Service - Privacy Policy