Cauchy Sequence: Understanding the Boundary Condition

In summary, the conversation discusses the concept of Cauchy sequences and provides examples to demonstrate how a sequence can be bounded but not Cauchy. The participants also discuss the definition of Cauchy sequences and provide explanations and counterexamples.
  • #1
steven187
176
0
hello all

I found this rather interesting
suppose that a sequence [tex]{x_{n}}[/tex] satisfies

[tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex] [tex] \forall n\epsilon N[/tex]

how couldn't the sequence [tex]{x_{n}}[/tex] not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

thanxs
 
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  • #2
An obvious one: just add those differences. Let xn be [tex]\sum_{i=1}^n \frac{1}{n}[/tex]. That series does not converge and so the sequence of partial sums is not Cauchy.
 
  • #3
I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

It looks like a Cauchy sequence to me based on the definition:

The sequence: [itex]\{x_n}\}[/itex] is a Cauchy sequence if given [itex]\epsilon[/itex] there exists N such that for all m,n[itex]\leq[/itex]N we have:

[tex]|x_m-x_n|<\epsilon[/tex]

For the sequence:

[tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex]


I take [itex]\epsilon=\frac{1}{N+1}[/itex];

Thus for all n>N:

[tex]|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon[/tex]
 
  • #4
Thus for all n>N:
[tex]|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon[/tex]

That's true; but look closely - is that really the same thing as

for all m,n[itex]\geq[/itex]N we have:[tex]
|x_m-x_n|<\epsilon[/tex]

?
 
  • #5
For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
You do remind me that just [tex]|x_{n+1}-x_n|< \epsilon[/tex] is not enough. You have to show that [tex]|x_n-x_m|< \epsilon[/tex] which is basically showing that the tail of the sum goes to 0.
 
  • #6
rachmaninoff said:
That's true; but look closely - is that really the same thing as



?


I suppose not.

[tex]|x_{n+1}-x_n|<\frac{1}{n+1}[/tex]

does not imply

[tex]|x_m-x_n|<\frac{1}{n+1}[/tex]

Still a little unclear. I'll work on it. Thanks guys.

Seriously, why do I even bother with differential equations anyway . . .
 
  • #7
Simply put: for given N, your

[tex] | x_m - x_n | [/tex],

[tex] \mbox{ as } m \rightarrow \infty [/tex],

is not guaranteed to be any better than the (divergent) harmonic series:

[tex] | x_m - x_n | < \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty [/tex]
 
  • #8
Alright, I'm interested in getting my math right.
Following Hall's example (a slight change):

Let:

[tex]x_n=\sum_{i=1}^{n+1} \frac{1}{i}[/tex]

thus:

[tex]|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}[/tex]

thus:

[tex]|x_{n+1}-x_n|=\frac{1}{n+2}<\frac{1}{n+1}[/tex]

However, the partial sums of this harmonic series [itex] \{x_n\}[/itex] do not converge and thus cannot be a Cauchy sequence.

Now, I got a pot of spagetti to make . . .
 
  • #9
hello all

nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i can't think of any other examples, do you know of any others?

steven
 
  • #10
hi guys i need bounded but not cauchy function can you help me asap?
 
  • #11
Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.
 

1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers where the terms gradually get closer and closer together as the sequence goes on. This means that for any small value, there is a point in the sequence where all subsequent terms are within that small distance from each other.

2. What is the significance of Cauchy sequences in mathematics?

Cauchy sequences are important in mathematics because they are used to define the concept of convergence in metric spaces. This allows us to determine whether a sequence of numbers converges to a specific limit or not.

3. How is the boundary condition related to Cauchy sequences?

The boundary condition in Cauchy sequences refers to the requirement that the sequence must have a limit or endpoint. This means that the terms in the sequence must eventually get close enough to each other to converge to a specific value.

4. What happens if the boundary condition is not satisfied in a Cauchy sequence?

If the boundary condition is not satisfied, then the sequence is not a Cauchy sequence. This means that the terms in the sequence do not get close enough to each other to converge to a limit, and the sequence is said to be divergent.

5. Can a Cauchy sequence have more than one limit?

No, a Cauchy sequence can only have one limit. This is because the sequence must converge to a specific value, and if it had more than one limit, it would not meet the criteria for a Cauchy sequence.

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