Proving a Sequence is Convergent

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SUMMARY

The discussion focuses on proving the convergence of a sequence {a_{n}}^{\infty}_{n=1} defined by the inequality |a_{n+1} - a_{n}| ≤ (1/2)|a_{n} - a_{n-1}| for n ≥ 2. The key conclusion is that demonstrating the sequence is Cauchy suffices for proving convergence. The participants confirm that |a_{m} - a_{n}| can be bounded by |a_{2} - a_{1|} multiplied by a geometric series, leading to the condition that for any ε > 0, there exists an N such that for all n, m > N, the sum of the series is less than ε/2|a_{2} - a_{1}|.

PREREQUISITES
  • Understanding of Cauchy sequences
  • Familiarity with geometric series
  • Knowledge of limits and convergence in real analysis
  • Basic manipulation of inequalities
NEXT STEPS
  • Study the properties of Cauchy sequences in real analysis
  • Learn about geometric series convergence criteria
  • Explore the formal definition of convergence in metric spaces
  • Investigate the implications of convergence on bounded sequences
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Students of real analysis, mathematicians interested in sequence convergence, and educators teaching concepts of limits and Cauchy sequences.

bobcat817
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Homework Statement


Let {a_{n}}^{\infty}_{n=1} be a sequence of real numbers that satisfies


|a_{n+1} - a_{n}| \leq \frac{1}{2}|a_{n} - a_{n-1}|

for all n\geq2


Homework Equations





The Attempt at a Solution



So, I know that it suffices to show that the sequence is Cauchy to prove this. And I can show that

|a_{m} - a_{n}| \leq \sum^{m-1}_{n}\frac{1}{2}^{k-1} * |a_{2} - a_{1}|

And that is about where I get lost. I'm very unclear on how to define the N in relation to the \epsilon.
 
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I will take your comment that you've shown

<br /> |a_m - a_n | \le |a_2 - a_1| \sum_{k=n}^m \left(\frac 1 2 \right)^k<br />

as correct - make sure it's correct.

Suppose \varepsilon &gt; 0 is given. Try taking N so large that n, m larger than N give

<br /> \sum_{k=n}^m \left(\frac 1 2 \right)^k &lt; \frac{\varepsilon}{2|x_2 - x_1|}<br />

This should work (why the 2 in my denom - to make sure the sequence terms differ by less than \varepsilon - it's a holdover from my student days, and I'm too old to change.)
 

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