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

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to answer true or false on whether or not the sequence ((-1)^n * n) tends toward both ±∞

    2. Relevant equations
    Uniqueness of Limits

    3. 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.
  2. jcsd
  3. Mar 16, 2012 #2
    Define "tends toward".

    If you mean converges to both ±∞, you can simply say no, since sequence cannot have two different limits.
  4. Mar 16, 2012 #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.
  5. Mar 16, 2012 #4


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    Hi Hodgey8806! :smile:

    I'd just prove (from the basic definition of limit) that it doesn't converge to +∞ :wink:
  6. Mar 16, 2012 #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.
    Last edited: Mar 16, 2012
  7. Mar 16, 2012 #6


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    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].
    Last edited by a moderator: Mar 16, 2012
  8. Mar 16, 2012 #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.
  9. Mar 16, 2012 #8


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    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?
  10. Mar 16, 2012 #9
    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.
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