Proving Sequence Convergence: ||s_n|-|L|| < epsilon

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SUMMARY

The discussion focuses on proving the inequality ||a|-|b|| <= |a-b| using the triangle inequality, which is a fundamental concept in real analysis. It also addresses the convergence of the sequence {|s_n|} to |L|, contingent upon the convergence of {s_n} to L. The Squeeze Theorem is highlighted as a method to demonstrate this convergence, emphasizing the importance of manipulating epsilon values in proofs.

PREREQUISITES
  • Understanding of the triangle inequality in real analysis
  • Familiarity with the Squeeze Theorem
  • Basic knowledge of sequence convergence
  • Proficiency in manipulating epsilon-delta definitions
NEXT STEPS
  • Study the properties of the triangle inequality in depth
  • Explore the Squeeze Theorem and its applications in proving limits
  • Learn about epsilon-delta definitions of limits and convergence
  • Practice problems involving sequence convergence and absolute values
USEFUL FOR

Students in real analysis, mathematics educators, and anyone seeking to deepen their understanding of sequence convergence and inequalities in mathematical proofs.

tarheelborn
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Homework Statement


For any a, b in R, show that ||a|-|b|| <= |a-b|. Then prove that {|s_n|} converges to |L| if {s_n} converges to L.

Homework Equations





The Attempt at a Solution


For the first part, ||a|-|b|| = |a-b| by the triangle inequality. For the second part, ||s_n|-0| < epsilon implies that |s_n -0| < epsilon, but I am not sure how to work that around to the L's.
 
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tarheelborn said:
For the first part, ||a|-|b|| = |a-b| by the triangle inequality.

What? For example:
\big|\,|2|-|-2|\,\big| = |2-(-2)|
?
 
Get the | |a| - |b| | \leq |a - b| right first. The second part just follows by the Squeeze Theorem, or more simply, just "shove in the epsilons" if you know what I mean.
 

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