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}|.

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  • Understanding of Cauchy sequences
  • Familiarity with geometric series
  • Knowledge of limits and convergence in real analysis
  • Basic manipulation of inequalities
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  • Learn about geometric series convergence criteria
  • Explore the formal definition of convergence in metric spaces
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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 {[tex]a_{n}[/tex]}[tex]^{\infty}_{n=1}[/tex] be a sequence of real numbers that satisfies


|[tex]a_{n+1}[/tex] - [tex]a_{n}[/tex]| [tex]\leq[/tex] [tex]\frac{1}{2}[/tex]|[tex]a_{n}[/tex] - [tex]a_{n-1}[/tex]|

for all n[tex]\geq[/tex]2


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

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

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

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

as correct - make sure it's correct.

Suppose [tex]\varepsilon > 0[/tex] is given. Try taking [tex]N[/tex] so large that [tex]n, m[/tex] larger than [tex]N[/tex] give

[tex] \sum_{k=n}^m \left(\frac 1 2 \right)^k < \frac{\varepsilon}{2|x_2 - x_1|}[/tex]

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

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