Arithmetic operations on sequences

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SUMMARY

The discussion focuses on proving that if the sequence {a_n} converges to a positive limit (a), then the sequence {sqrt(a_n)} also converges to sqrt(a). The key approach involves manipulating the expression |(a_n - a) / (sqrt(a_n) + sqrt(a)| to demonstrate convergence. Participants suggest using the conjugate to facilitate the proof, emphasizing the importance of bounding the difference between the terms to establish the limit rigorously.

PREREQUISITES
  • Understanding of limits and convergence in sequences
  • Familiarity with the properties of square roots
  • Knowledge of epsilon-delta definitions of limits
  • Basic algebraic manipulation techniques
NEXT STEPS
  • Study the epsilon-delta definition of limits in more depth
  • Learn about the properties of convergent sequences
  • Explore the use of conjugates in limit proofs
  • Practice problems involving convergence of sequences and their transformations
USEFUL FOR

Students in calculus or real analysis, mathematicians focusing on sequence convergence, and educators teaching limit concepts in higher mathematics.

dssmith
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Homework Statement



If the sequence {a_n} n=1 to infinity converges to (a) with a_n >0 show {sqrt(a_n)}
converges to sqrt(a)

Homework Equations



hint: conjigate first

The Attempt at a Solution



abs[ (a_n-a) / (sqrt(a_n)+sqrt(a) ) ] < epsilon

i do not own LATEX, yet.
 
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You know that for some M, and for all n >= M, there is a positive epsilon such that |an - a| < epsilon.

I would approach this by factoring |an - a| into [itex]|\sqrt{a_n} - \sqrt{a}||\sqrt{a_n} + \sqrt{a}|[/itex]. Then maybe you can replace the second factor by something clever.
 
Clever like WHAT?? Waaaaah :'(
 

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