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(Real Analysis)For example, given a sequence (𝑥𝑛) so that there are A>0, C ∈(0,1) and K ∈ â„• so that|𝑥𝑛+1âˆ...

Question

(Real Analysis)For example, given a sequence (𝑥𝑛) so that there are A>0, C ∈(0,1) and K ∈ ℕ so that|𝑥𝑛+1−𝑥𝑛|≤𝐴𝐶𝑛,∀𝑛≥𝐾.Prove (𝑥𝑛) is a Cauchy Sequence.

(Real Analysis) For example, given a sequence (𝑥𝑛) so that there are A>0, C ∈ (0,1) and K ∈ ℕ so that |𝑥𝑛+1−𝑥𝑛|≤𝐴𝐶𝑛,∀𝑛≥𝐾. Prove (𝑥𝑛) is a Cauchy Sequence.



Answers

A sequence $a_{1}, a_{2}, \cdots$ is called a Cauchy sequence $^{\dagger}$ if for each $\epsilon>0$ there exists an index $k$ such that $\left|a_{n}-a_{m}\right|<\epsilon \quad$ for all $m, n \geq k$ Show that every convergent sequence is a Cauchy sequence. It is also true that every Cauchy sequence is convergent, but that is more difficult to prove.

Hello. So recall a sequence esteban is set to be a koshi sequence. If and only if um for every epsilon grading zero. So for all epsilon greater than zero um there is going to exist a positive integer end. So there exists a positive integer. And um such that the absolute value of S sub n minus S sub M is going to be less than epsilon. And this is for all and and M. That is going to be greater than our capital. And okay so we're gonna show that every convergent sequence is koshi. Well let? S see Ben um converge to L. So say let's have that S C. Ben is going to converge to L. Okay, this means that for all epsilon greater than zero there is going to exist um an M. That's an element of the positive integers. Um Such that we have the absolute value of S sub n minus L. Is going to be less than epsilon over to for all N greater than or equal to M. So if we have that people and Q are greater than or equal to M then we have that the absolute value of a S a p minus L. Is going to be less than epsilon over two. And we have that the absolute value of S sub q minus L is going to be less than epsilon over two. So we have that the absolute value of S A p minus S sub Q is going to be equal to well the absolute value of S A p minus L plus l minus S P. Um which is going to be less than or equal to the absolute value of S p minus L plus the absolute value of S sub q minus L which is going to be less than excellent over two plus Excellent over two, which is going to be less than epsilon. So that is for every epsilon greater than zero. There exists a positive end such that the absolute value of S a p minus s of Q is less than epsilon for all P and q greater than N. So therefore we have that are sequence as sub N is going to be a koshi a koshi sequence. All right. Take care.


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