Real analysis proof with sequences

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SUMMARY

The discussion centers on proving that the limit of a sequence Sn in R approaches 0 if and only if the limit of the absolute value of the sequence, |Sn|, approaches 0. The proof involves demonstrating that if lim Sn = 0, then for sufficiently large n, |Sn| must also be less than any given epsilon. Conversely, if lim |Sn| = 0, a similar argument shows that Sn must also approach 0. This establishes that the two limits are equivalent without circular reasoning.

PREREQUISITES
  • Understanding of limits in real analysis
  • Familiarity with epsilon-delta definitions of limits
  • Knowledge of sequences and their properties
  • Basic proof techniques in mathematical analysis
NEXT STEPS
  • Study the epsilon-delta definition of limits in detail
  • Explore the properties of convergent sequences in real analysis
  • Learn about the relationship between sequences and their absolute values
  • Practice constructing proofs in real analysis
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Students of real analysis, mathematicians focusing on sequences and limits, and anyone seeking to strengthen their proof-writing skills in mathematical contexts.

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Homework Statement


Let Sn be a sequence in R

Prove lim Sn= = 0 if and only if lim abs(Sn) = 0

Homework Equations



none

The Attempt at a Solution



I think this is someone ciruclar logic and that is why I am stuck

Assume lim Sn = 0, thus for n > N implies |Sn| < epsilon or -epsilon < Sn < epsilon. Since Sn and -Sn are less than epsilon |Sn| < Epsilon for sufficently large n.
Now assume lim |Sn|= 0 so ||Sn||< epsilon and using a similar argument show Sn < Epsilon and complete the proof. IS that the way to go?
 
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It's not circular, it's just that proving Sn->0 and |Sn|->0 are pretty much the same thing. Yes, that's the way to go.
 

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