Proof of Limit of (Tn*Sn)=0 Using Definition

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SUMMARY

The discussion focuses on proving that the limit of the product of a bounded sequence (Tn) and a sequence (Sn) converging to zero is also zero, using the definition of a limit. The user is guided to utilize the completeness axiom of real numbers and the properties of bounded sequences. Key steps include establishing an upper bound M for |Tn| and demonstrating that both -MSn and MSn converge to zero, leading to the conclusion that SnTn approaches zero as well.

PREREQUISITES
  • Understanding of limits of sequences in real analysis
  • Familiarity with the completeness axiom of real numbers
  • Knowledge of bounded sequences and their properties
  • Ability to manipulate inequalities involving sequences
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  • Explore the completeness axiom of real numbers and its implications
  • Learn about bounded sequences and their convergence properties
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Students and educators in mathematics, particularly those studying real analysis and limits, as well as anyone looking to strengthen their understanding of sequence convergence and proof techniques.

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Homework Statement



Let (Tn) be a bounded sequence and let (Sn) be a sequence whose limit = 0. PRove that the limit of (Tn*Sn) = 0. I must complete the proof using only the definition of a limit of a sequence

Homework Equations


Let Sn be a sequence from N->R with a limit of s
For all epsilon > 0 there exists an n > N, such that |Sn - s| < epsion
The suprenum of a sequence is the least upper bound

The Attempt at a Solution



I am used to the numerical proofs showing limits I am not comfortable with manipulating these more abstract ones. I cannot solve for N so I am stuck but here is what I am working off of...


Since Tn is a bounded sequence, by the completeness axiom of R I know that a suprenum exisit and I call it U

Since Sn is a sequence that converges I know for epsilon > 0 there exist an n > N(For s) such that |Sn|< epsilon

I need to show that there is an N for any epsilon that implies |Sn*Tn - 0*U| < epsilon

I was debating proving that the Tn limit is the sup T then provoing the multiplaction of limits rule but that does not sound right there should be a direct method I need some hints on how to start this proof many thanks
 
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You cannot use that "Tn limit is the sup T then proving the multiplaction of limits rule" because you do not know that Tn has a limit. All you know about Tn is that it is bounded. For example, Tn= (-1)n is bounded but does not converge.

Since Tn is bounded, let M be an upper bound for |Tn|. Now you know that -MS_n\le T_nS_n\le MS_n. Combine that with the fact that Sn converges to 0.
 
Thanks Halls for the response

Just so I got to close up the proof. I can say for a large N per the hypothesis Sn goes to zero so
MSn and -MSn go to zero.
It follows that SnTn goes to zero
and since epsilon > 0 MSn < epsilon.
am I done or is it more involved?

Am I right in saying for these more abstract sequence that you cannot really prove via the definition of a limit of a sequence. The goal is to show that something is less for any given epsilon?
 

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