I just have a question about Uniqueness of Limits with divergent sequences.

In summary: Yes, the proof is valid in this case. The theorem on uniqueness of limits states that a sequence cannot have 2 distinct limits. In this case, the sequence has a limit of -∞ and a limit of +∞. Since the sequence has 2 distinct limits, the theorem on uniqueness of limits is valid.
  • #1
Hodgey8806
145
3

Homework Statement


I'm supposed to answer true or false on whether or not the sequence ((-1)^n * n) tends toward both ±∞


Homework Equations


Uniqueness of Limits


The Attempt at a Solution


I did prove it another way, but I would think that uniqueness of limits (as a definition available for use) is enough to disprove this statement.
 
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  • #2
Define "tends toward".

If you mean converges to both ±∞, you can simply say no, since sequence cannot have two different limits.
 
  • #3
I understand that, but the teacher asks me to prove that. Since we previously proved uniqueness of a limit, I'm thinking we can just expand it to apply to divergent sequences.

Tends toward is to say that after a certain natural number K, any n>=K implies that Xn > a FOR ALL a in R--this is the definition of tends toward infinity. It ultimately heads toward infinity.
 
  • #4
Hi Hodgey8806! :smile:

I'd just prove (from the basic definition of limit) that it doesn't converge to +∞ :wink:
 
  • #5
Hello :) I did prove it another way. But that involves the fact that if a sequence tends toward infinity, it has a lower bound hence it doesn't tend toward negative infinity. Similarly, if it tends toward negative infinity, then it can't tend toward positive infinity.

But I would think Uniqueness would suffice.

The next proof was to prove the negation true that it does NOT tend toward negative infinity nor positive infinity.
 
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  • #6
The sequence [itex]a_n= (-1)^n n[/itex] does NOT converge, even in the sense of "to [itex]-\infty[/itex]" and "to [itex]-\infty[/itex]". However, it has two subsequences to [itex]diverge[/itex] to those values: [itex]a_n[/itex], for n even, diverges to [itex]+\infty[/itex] while [itex]a_n[/itex], for n odd, diverges to [itex]-\infty[/itex].
 
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  • #7
Thank you. I'm aware that it does not converge in that sense. But I have to formally explain why that is the case. Hence, I'm using bounds to show that if you settle with it diverging to one of the infinities, you're forced to agree that it doesn't diverge to the other (by the idea of lower bounds).

This implies that the negation is true.

I'm aware that the sub-sequences can be forced to diverge to the desired limits. But the question is about the sequence given. What it is proving is that if this is false, then the negation is actually true. But I just need to formally show that it is false.

Instead of writing the proof that I did on here, I'd rather just state a known definition that Uniqueness of Limits tells us that a sequence cannot have 2 distinct limits--However, I'm not quite sure that applies fully to divergent sequences.
 
  • #8
Hodgey8806 said:

Homework Statement


I'm supposed to answer true or false on whether or not the sequence ((-1)^n * n) tends toward both ±∞

Homework Equations


Uniqueness of Limits

The Attempt at a Solution


I did prove it another way, but I would think that uniqueness of limits (as a definition available for use) is enough to disprove this statement.
Look at the proof for the theorem on uniqueness of limits. Is that proof valid in the case where the limit +∞ or -∞ rather than the limit being a finite value?
 
  • #9
SammyS said:
Look at the proof for the theorem on uniqueness of limits. Is that proof valid in the case where the limit +∞ or -∞ rather than the limit being a finite value?

That's what my question is. Is it expandable to that sense? My book only gives it in the section concerning finite limits. However, I have proved this in the manner concerning bounds. I'm just not sure that the definition of limit uniqueness is actually allowable in this sense.
 

1. What is a divergent sequence?

A divergent sequence is a sequence of numbers that does not converge to a specific limit. This means that as the sequence continues, the numbers will either increase or decrease without approaching a fixed value.

2. How do you find the limit of a divergent sequence?

Since a divergent sequence does not have a limit, it is not possible to find the exact value of the limit. However, we can still analyze the behavior of the sequence and determine if it is approaching positive or negative infinity, or if it is oscillating between different values.

3. Can a divergent sequence have a limit?

No, by definition, a divergent sequence does not have a limit. In order for a sequence to have a limit, it must converge to a specific value.

4. How is the uniqueness of limits related to divergent sequences?

The uniqueness of limits states that a sequence can only converge to one specific limit. Since a divergent sequence does not have a limit, this concept does not apply to it.

5. Can the uniqueness of limits be violated with divergent sequences?

No, the uniqueness of limits cannot be violated with divergent sequences because they do not have a limit to begin with. Therefore, there is no limit to be violated.

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