I know that a contractive sequence is a Cauchy sequence. Is the converse true? Does a sequence being Cauchy implies that it is contractive, if the sequence is defined over the set of real numbers? If not, please provide a mathematical explanation/proof supported with a counter-example.
Is every Cauchy sequence contractive?
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real-analysis
sequences-and-series
cauchy-sequences
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1What is the precise definition of contractive sequence? – 2017-02-13
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0A sequence which satisfies the contractive property $|x_{n+1}-x_n|<=a*|x_{n}-x_{n-1}|$ for all $n$ such that $0
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0We can replace any finite number of members of a Cauchy sequence with arbitrary values and it will still be a Cauchy sequence.... Also, if $(x_n)_n$ is Cauchy with $x_n\ne x_{n+1}$ for all $n$ then $(y_n)_n$ is Cauchy where $y_{2n}=y_{2n+1}=x_n.$ – 2017-02-13
2 Answers
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No. For example consider $a_n = 1/n$.
Then $\frac{|a_{n+2}-a_{n+1}|}{|a_{n+1}-a_n|} = \frac{n(n+1)}{(n+1)(n+2)} \to 1$ as $n \to \infty$.
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I assume that the underlying metric space is $\mathbb{R}$.
Since every Cauchy sequence is convergent and since there exist non-contractive convergent sequences, I guess that the answer is "No".