Proving that a sequence is confined to a certain interval

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Discussion Overview

The discussion revolves around proving that a given sequence is eventually confined to the interval [0, a] for any number a > 1. The sequence in question is defined as alternating and involves exploring its behavior as n increases. Participants are working through the proof and addressing challenges related to the alternating nature of the sequence.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the sequence and expresses confusion about proving its confinement to the interval [0, a], particularly due to its alternating nature.
  • Another participant suggests that for sufficiently large n, the sequence can be shown to be less than a by choosing an appropriate N based on a given a.
  • There is a discussion about the behavior of the sequence for even and odd powers, with one participant questioning the necessity of considering both cases in the proof.
  • Clarification is provided regarding the cancellation of signs for odd powers, and the implication that even powers will always yield values less than 1.
  • Participants express enthusiasm and a sense of accomplishment in engaging with the mathematical problem.

Areas of Agreement / Disagreement

Participants generally agree on the approach to proving the sequence's confinement but have not reached a consensus on the necessity of addressing both even and odd powers in detail. Some uncertainty remains regarding the implications of the alternating nature of the sequence.

Contextual Notes

Participants are navigating the complexities of alternating sequences and their convergence properties, with some assumptions about the behavior of the sequence as n increases remaining implicit.

Who May Find This Useful

This discussion may be useful for students studying sequences and series, particularly those dealing with alternating sequences and convergence proofs in a mathematical context.

Claire84
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In class we were given the following sequence-

(0,1 1/2, 3/4, 1 1/8, 15/16, 1 1/32 ...)

So it's alternating about zero. We're told then that given any number a>1 it s true to say that the sequence (xn) is eventually confined t [0,a]

We now have to prove this...

Inworked out the formula for the sequence t be 1- (-1/2)^(n-1) but I'm not sure if that was necessary or not.

Then I set about getting the poof by working backwards to get an idea of what it was going to be like, but I'm confuse because it's an alternating sequence and we haven't done any like that yet. I asked one o the phd guys about it and they looked at it for half an hour and couldn't do it, but it's only a level 1 problem so can someone else give some feedback on it? Here's what I've come up with so far for my idea for the proof...

Since we have n>no

1- (-1/2)n-1 > 1-(-1/2)^(no-1) for even powers...

so we have a> 1- (1/2)^(n-1)

(-1/2)^(n-1)>1-a

since it's an even power you could then have-

1/2^(n-1)> 1-a

2/(2^n)>1-a

so you could then start the proof with a number (1-a)/2>0

Hoqwever, I don't know what to do for uneven powers because then wouldn't 1-(-1/2)^(n-1) < 1- (-1/2)^(no -1)
 
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You're on the right track.

Let a_n be the sequence 1-(-1/2)^n as given

we must show that for all a>1, a_n<a for all n sufficiently large.

Given a, let N be chosen such that 2^(-N)<a-1, which can be done (by eg archimedes' axiom)

|a_n| <= 1+(1/2)^n for all n (ie we can ignore the minus sign) and if n>N then |a_n| < 1+(1/2)^N < a
 
Last edited:
|a_n| <= 1+(1/2)^n for all n (ie we can ignore the minus sign)

is this because this would have bee like the odd power and therefore the 2 minuses cancel out? And then you don't really need to think about the even power thing I dscribed because it's obviously going to be less than one? Would you still need to go through the even power situation? Sorry, this is all a bit new to me but I'm glad I actually managed to do something relatively logical in my first post.:smile:
 
yep, (-x)^n = -(x^n) if n is odd, so the signs cancel. you at least need to state that the n even case is clearly always less than 1
 
Aha, thanks for that! Suddenly feeling very brainy! *cough* ell a least it shows I'm enthusiatsic that I'm doing Maths on St.Patrick's day!
 

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