Bounded sequence that diverges, convergent subsequence

So, the sequence (s1, s1, s1, ...) is a subsequence of (a_{n}).In summary, to show that (sn) has at least two convergent subsequences with different limits, you can use the Bolzano-Weierstrass theorem and show that there are two sequences that converge to the infima and suprema and that these limits are not equal. Alternatively, you can show that there is a subsequence of (sn) that does not converge to the same limit as the Bolzano-Weierstrass sequence, and then show that this subsequence has a convergence subsequence with a different limit.
  • #1
miren324
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Homework Statement


Let (sn) be a sequence in R that is bounded but diverges. Show that (sn) has (at least) two convergent subsequences, the limits of which are different.


Homework Equations





The Attempt at a Solution


I know that a convergent subsequence exists by Bolzano-Weierstrass. I'm having trouble showing that multiple convergent subsequences exist, and how to show that they would have different limits. I know that (sn) will probably look something like (-1)^n (so it will be a sequence that neither increases or decreases, but rather most likely "bounces around"), and if I had an actual sequence to prove this on I could probably do it, but to prove it in general is giving me some difficulty.

The only thought I had was that maybe defining two subsequences of (sn) that have only one element, i.e. (an) = {s1} and (bn) = {s2}. I'm not sure this would work, however, since I am not sure that an infinite constant sequence, such as (an), would be a subsequence of (sn), seeing as how, (s1) would occur much fewer times in (sn) than it would in (an) (for example: (sn) = sin(n), (an) = sin(1), then (an) = {sin1, sin1, sin1, sin1, ...} and (sn) = {sin1, sin2, sin3, sin4, ...}).

Any help would be greatly appreciated.
 
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  • #2
The approach to this proof depends on what theorems you are allowed to use. For instance, if you let S be the set of subsequential limits of [tex]s_{n}[/tex] then there is a theorem which states

[tex]\sup S = \lim\sup s_{n}\mbox{ and }\inf S = \lim\inf s_{n}[/tex]

and all you need to do is prove [tex]\lim\sup s_{n}\neq\lim\inf s_{n}[/tex]

which will be easy since you know that [tex]s_{n}[/tex] isn't convergent.
 
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  • #3
None of my proof are correct. Only one sequence converges the other I made up did not. But we need both to converge.
 
  • #4
The correct solution is to show that there are two sequences that converge to the infima and suprema and infima is not equal to the suprema.

Another correct approach is to show that there is a subsequence of Sn that does not converge to the same limit as your bolzano weierstrass sequence and then show that that subsequence has a convergence subsequence that converges also to a different limit than your original BW sequence.
 
  • #5
The only thought I had was that maybe defining two subsequences of (sn) that have only one element, i.e. (an) = {s1} and (bn) = {s2}. I'm not sure this would work, however, since I am not sure that an infinite constant sequence, such as (an), would be a subsequence of (sn), seeing as how, (s1) would occur much fewer times in (sn) than it would in (an) (for example: (sn) = sin(n), (an) = sin(1), then (an) = {sin1, sin1, sin1, sin1, ...} and (sn) = {sin1, sin2, sin3, sin4, ...}).

You can do an infinite constant subsequence, just not the way illustrate in your sin example. You can't take (an)=(s1) as a subsequence simply because of the definition of a subsequence; however, you can defined
[tex]a_{n}=sin(2\pi n)[/tex] and then your sequence is the constant 0. This you are allowed to do this because sin(n) takes on the value of 0 infinitely many times.
 
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FAQ: Bounded sequence that diverges, convergent subsequence

What is a bounded sequence that diverges?

A bounded sequence is a sequence of numbers where the values do not exceed a certain limit. A sequence that diverges is one that does not have a limit and its terms continue to grow or decrease indefinitely.

How can you determine if a sequence is bounded?

To determine if a sequence is bounded, you can look at its terms and see if they are increasing or decreasing without bound. If the terms do not have a limit and continue to grow or decrease, the sequence is not bounded.

3. What is a convergent subsequence?

A convergent subsequence is a subset of a sequence that has a limit. This means that as the terms of the subsequence approach infinity, they will eventually converge to a single value.

4. How can you identify a convergent subsequence in a bounded sequence that diverges?

In a bounded sequence that diverges, there may be subsets of terms that have a limit and converge to a single value. These subsets are the convergent subsequences. To identify them, you can look for patterns in the terms or use mathematical techniques such as the Bolzano-Weierstrass theorem.

5. Why is understanding bounded sequences that diverge and their convergent subsequences important in science?

Bounded sequences that diverge and their convergent subsequences are important in science because they help us understand the behavior of systems with changing values. In many scientific fields, such as physics and biology, understanding the limits and patterns of sequences is crucial in making predictions and drawing conclusions from data.

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