About square summable sequences space

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SUMMARY

The discussion centers on disproving the statement that a sequence \( (x_n) \in \ell^2 \) is a Cauchy sequence if \( \lim_{n \to \infty} d(x_n, x_{n+1}) = 0 \). The key takeaway is that the sequence \( x_n = \left(1, \frac{1}{2}, \ldots, \frac{1}{\sqrt{n}}\right) \) is square summable, yet does not satisfy the Cauchy condition as defined by \( \exists n_0 \; \forall p,q > n_0 \; d(x_p, x_q) \rightarrow 0 \). The counterexample provided, \( \sum_{n=1}^\infty \frac{1}{n} \), illustrates that convergence does not imply Cauchy behavior in this context.

PREREQUISITES
  • Understanding of Cauchy sequences
  • Familiarity with square summable sequences in the context of functional analysis
  • Knowledge of distance metrics in sequence spaces
  • Basic concepts of convergence and divergence in series
NEXT STEPS
  • Study the properties of Cauchy sequences in metric spaces
  • Explore the implications of the Cauchy criterion in functional analysis
  • Investigate the characteristics of \( \ell^2 \) spaces and their convergence properties
  • Learn about counterexamples in analysis, particularly in relation to series convergence
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Mathematicians, students of functional analysis, and anyone interested in the properties of sequences and convergence in mathematical analysis.

antiņš
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First, I'm sorry for my bad english.

Homework Statement


I need to disprove:
(x_n) \in \ell^2 is a Cauchy sequence, if \displaystyle \lim_{x \to \infty} d(x_n, x_{n+1})=0.

Homework Equations


Ok, sequence is Cauchy sequence if \exists n_0 \; \forall p,q>0 \; d(x_p,x_q) \rightarrow 0


The Attempt at a Solution


Has someone idea about this? I tried 1/ln(x) and many examples like this one, but all this are wrong.
 
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antiņš said:
First, I'm sorry for my bad english.

Homework Statement


I need to disprove:
(x_n) \in \ell^2 is a Cauchy sequence, if \displaystyle \lim_{x \to \infty} d(x_n, x_{n+1})=0.

Homework Equations


Ok, sequence is Cauchy sequence if \exists n_0 \; \forall p,q>0 \; d(x_p,x_q) \rightarrow 0
You mean \forall p,q> n_0

The Attempt at a Solution


Has someone idea about this? I tried 1/ln(x) and many examples like this one, but all this are wrong.
Here's a hint: \sum_{n=1}^\infty\frac{1}{n} does not converge.

Your English is excellent. (Well, except for not capitalizing "English"!)
 
Of course, x_n=(1, \frac{1}{2}, ... , \frac{1}{\sqrt{n}}) is what I'm looking for. And this sequence is square summable because x_n is finite.

Thanks! ^_^
 
Last edited:
antiņš said:
First, I'm sorry for my bad english.

No worries mate, Latvians speak English well.
 

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