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 ```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``` 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```