Proving Convergence with Cauchy Sequence Method

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Homework Help Overview

The discussion revolves around proving the convergence of a sequence defined by the condition |s_{n+1} - s_n| < 2^{-n} for all natural numbers n. The original poster is exploring the approach of demonstrating that the sequence is a Cauchy sequence as a means to establish its convergence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the validity of using the Cauchy sequence definition to prove convergence. There are suggestions to utilize the triangle inequality and consider the implications of the sequence's properties. Some participants emphasize the need to show that the difference |a_m - a_n| approaches zero as m and n increase independently.

Discussion Status

The conversation is ongoing, with participants providing insights into the Cauchy sequence approach. There is a focus on the necessity of demonstrating the general case through induction and the application of the sequence's defining property. Multiple lines of reasoning are being explored without a clear consensus yet.

Contextual Notes

There is an emphasis on the requirement that m and n can be any natural numbers, not just consecutive indices. The original poster expresses uncertainty about the approach, indicating a need for clarification on the method to prove convergence.

JamesF
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Hello all. I'm having trouble on the following homework problem. It seems like it should be easy, but I'm just now sure how to approach it

Homework Statement



Let [tex](s_n)[/tex] be a sequence st [tex]|s_{n+1} - s_n | < 2^{-n}, \forall n \in \mathbb{N}[/tex]

show that [tex](s_n)[/tex] converges


The Attempt at a Solution



well I thought the easiest way to prove it would be to show it's a Cauchy sequence and therefore convergent, but perhaps that's the wrong approach. What's a good starting point for a problem like this?
 
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A good starting point is to show it's Cauchy. But you already knew that. Just do it. Cauchy means |s_n-s_m|<epsilon for n and m greater than N. The maximum difference between s_n and s_m involves summing a lot of large powers of 1/2. Use the triangle inequality.
 
To show that a sequence is Cauchy you must show that [itex]|a_m- a_n|[/itex] goes to 0 as m and n go to infinity independently (in particular, you cannot assume that m= n+1).

But [itex]|a_{n+2}- a_n|\le |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/itex], [itex]|a_{n+3}- a_n|\le |a_{n+3}- a_{n+2}|+ |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/itex], etc.

You can use the given property on each of those and use induction to show the general case.
 
HallsofIvy said:
To show that a sequence is Cauchy you must show that [itex]|a_m- a_n|[/itex] goes to 0 as m and n go to infinity independently (in particular, you cannot assume that m= n+1).

But
[tex]|a_{n+2}- a_n|\le |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/tex]
[tex]|a_{n+3}- a_n|\le |a_{n+3}- a_{n+2}|+ |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/tex], etc.

You can use the given property on each of those and use induction to show the general case.
asaaaa
 

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