Sequences ratio test, intro to real analysis

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SUMMARY

The discussion centers on proving that a sequence of positive real numbers, defined by the limit condition lim(xn+1 / xn) = L > 1, is not bounded and therefore not convergent. The proof utilizes the definition of convergence, which states that for every epsilon > 0, there exists a natural number K such that for all n ≥ K, |xn - x| < epsilon. By applying epsilon = (L-1)/2, it is established that for n > N, x_n+1 grows at least as fast as a power series with a ratio greater than 1, confirming the sequence's unbounded nature.

PREREQUISITES
  • Understanding of real analysis concepts, particularly limits and convergence.
  • Familiarity with sequences and their properties in mathematical analysis.
  • Knowledge of the epsilon-delta definition of limits and convergence.
  • Basic skills in constructing mathematical proofs.
NEXT STEPS
  • Study the properties of sequences in real analysis, focusing on boundedness and convergence.
  • Learn about the ratio test and its applications in determining convergence of series.
  • Explore the implications of the epsilon-delta definition in more complex proofs.
  • Investigate power series and their convergence criteria in mathematical analysis.
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Students of real analysis, mathematicians focusing on sequences and series, and anyone interested in understanding the foundations of convergence in mathematical proofs.

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Homework Statement



Let X = (xn) be a sequence of positive real numbers such that lim(xn+1 / xn) = L > 1.


Show that X is not a bounded sqeuence and hence is not convergent.


Homework Equations


Definition of convergence states that for every epsilon > 0 there exist some natural number K such that for all n > = K, |xn - x| < epsilon, then the squence converges to x.


The Attempt at a Solution



This is a proof so there really isn't too much to say here. I have looked at the definition of convergence and I see that I can get something like xn+1 < xn(L+e) where e is epsilon, but I do not see any way to produce an upper bound from that. I also know that the previous statement is true for all e, but that still does not seem to get me anywhere. I know I need to use the fact that L > 1, but I don't see how at this point.


Any hints, subtle or not, are welcome.

Thanks,

The Geekster
 
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Use epsilon=(L-1)/2. That means for some N, for all n>N. x_n+1>((L+1)/2)*x_n. (L+1)/2>1. So for n>N x_n increases at least as fast as a power series with a ratio > 1.
 

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