Recursive sequence convergence

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SUMMARY

The recursive sequence defined by x_n = (1/3)x_{n-1} + (2/3)x_{n-2} converges for arbitrary real numbers x_1 and x_2. The sequence is shown to be contractive, which confirms its convergence based on the properties of Cauchy sequences. The next step involves determining the limit of the sequence (x_n), which can be approached by analyzing the fixed points of the recursive formula.

PREREQUISITES
  • Understanding of Cauchy sequences
  • Familiarity with recursive sequences
  • Knowledge of contractive mappings
  • Basic calculus concepts related to limits
NEXT STEPS
  • Study the properties of Cauchy sequences in detail
  • Learn about fixed point theorems and their applications
  • Explore the concept of contractive mappings in metric spaces
  • Investigate methods for finding limits of recursive sequences
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Mathematics students, educators, and anyone interested in the convergence of recursive sequences and Cauchy sequences.

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Homework Statement



Let [tex]x_1 < x_2[/tex] be arbitrary real numbers and let [tex]x_n :=\frac{1}{3}x_{n-1} + \frac{2}{3}x_{n-2}[/tex]. Prove the sequence [tex](x_n)[/tex] converges.

Homework Equations



Since this problem comes from the section on Cauchy sequences, I assume we will need to show [tex](x_n)[/tex] is a Cauchy sequence. I'm not so well-versed in working with the recursive sequences especially with arbitrary initial values.

Any advice on getting started?
 
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would this be a valid solution? it looks like i can show the sequence is contractive.

[tex]|x_{n+1}-x_n| = |\frac{1}{3}x_n + 2 x_{n-1} - x_n | = \frac{2}{3}|x_{n-1} - x_n|[/tex]

Thus [tex](x_n)[/tex] is contractive, so it is convergent.
 
Last edited:
OK. so now how do we go about finding the limit of [tex](x_n)[/tex]?
 

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