Bounded sequence as convergent

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Discussion Overview

The discussion revolves around the properties of bounded sequences in relation to convergence, specifically addressing whether all bounded sequences must be convergent. Participants explore examples and definitions related to this concept.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that not all bounded sequences are convergent, citing the sequence Xn=(-1)^n as an example.
  • One participant suggests that to demonstrate divergence, it is sufficient to find two subsequences of Xn that converge to different limits.
  • Another participant expresses frustration and suggests revisiting definitions related to convergence.
  • There is a reiteration that if a sequence converges, it must be bounded, but being bounded does not guarantee convergence.
  • Participants discuss the negation of convergence and its implications for proving divergence, emphasizing the need to check inequalities related to convergence definitions.
  • One participant proposes that to prove a specific sequence is not convergent, one can exhibit two subsequences converging to different limits, specifically mentioning the subsequences of Xn that converge to 1 and -1.

Areas of Agreement / Disagreement

Participants generally agree that bounded sequences can be divergent, but there is no consensus on the best approach to prove this, leading to multiple competing views and methods being discussed.

Contextual Notes

Some discussions involve assumptions about definitions of convergence and the conditions under which sequences are considered bounded or convergent. The mathematical steps and inequalities discussed remain unresolved.

electronic engineer
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Some rule says that not all bounded sequence must be convergent sequence , one example is the sequence with general bound:
Xn=(-1)^n

could anyone help?!
thanks in advance!
 
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What is it that you want help with? To show that x_n is divergent, it suffices to find two subsequences of x_n which converge to different numbers.
 
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Ooh, erm, let me think... how about looking at the bloody definitions? Yes, I am tired, so delete as applicable, but quite frankly, I've seen enough of this for one day.
 
electronic engineer said:
Some rule says that not all bounded sequence must be convergent sequence , one example is the sequence with general bound:
Xn=(-1)^n
could anyone help?!
thanks in advance!

if a sequence converges then it's bounded, but if it's bounded it might not converge as that example shows.
 
fourier jr said:
if a sequence converges then it's bounded, but if it's bounded it might not converge as that example shows.

you're right, the question is how to prove that,could you help?!
 
What is the negation of: x_n converges to x? Show that this is satisfied. Of course if x_n converges to x then all subsequences of x_n converge to x as well, maknig for an easy proof that it doesn't converge. If you're not used to working out the negations of propositions then say so.
 
electronic engineer said:
you're right, the question is how to prove that,could you help?!
suppose that the sequence {a_n} converges. then a_n --> L for some finite L. now look at the definition of convergence & see that for ANY epsilon (yadda yadda)... for n>N but to make things easier just choose epsilon=1 & if n>N get 2 inequalities |a_n| < |a_n - L| + |L| < 1 + |L| (i think that's right, you check it). if n<N then |a_n| < something else.

so for any n>0, |a_n| < M where M=max{a certain set of numbers}. end of proof
 
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electronic engineer said:
you're right, the question is how to prove that,could you help?!
Prove what? That there exist bounded sequences that are not convergent? You do that by exhibiting one, just as you did.

Prove that that particular sequence is not convergent? Just as Muzza said in the very first response to your post: show that there exist two different subsequences that converge to two different limits- in this case the subsequence with n even: 1, 1, 1, ... converges to 1, the subsequence with n odd: -1, -1, -1, ... converges to -1. Since arbitrarily far into the sequnce there exist numbers arbitrarily close to 1 and numbers arbitrarily close to -1, taking [itex]\epsilon= 1/3[/itex] will show that no number can be the limit of the entire sequence.
 
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