Proof of limsup(anbn)<=limsup(an)limsup(bn) and Example

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Homework Help Overview

The problem involves proving a relationship between the limit superior of the product of two bounded positive sequences and the limit superiors of the individual sequences. The original poster seeks to establish that limsup(anbn) is less than or equal to limsup(an) times limsup(bn) and to provide an example illustrating that equality does not hold in general.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Some participants discuss the properties of limit superior and attempt to express the sequences in terms of their limit superiors. Others question the assumptions regarding the boundedness and positivity of the sequences.

Discussion Status

The discussion is ongoing, with participants exploring the relationship between the sequences and their limit superiors. Guidance has been offered to clarify the initial steps needed to approach the proof, but no consensus has been reached on the method or final outcome.

Contextual Notes

There is an emphasis on the requirement for participants to demonstrate their own attempts at solving the problem to maintain the thread's validity. The original poster has provided a specific example of sequences, but further details are not yet discussed.

stukbv
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Homework Statement


if (an) and (bn) are bounded positive sequences prove that
limsup(anbn)<= limsup(an)limsup(bn)
and give an example to show there is not equality in general
 
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If you do not show at least some attempt to do this problem yourself, this thread will be deleted.
 
all i know is, if we let l = limsupan and m = limsupbn ,
then we can say that there exists an n such that
l-e < an < l+e
m-e<bn < l + e

so |anbn| < (l+e)(m+e)
...
 
Hi stukbv! :smile:

Can you first show that

\sup{(a_nb_n)}\leq \sup{a_n}\sup{b_n}
 

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