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Doubt regarding the sequence (-1)^n

  1. Aug 24, 2012 #1
    Hello everyone!

    I was wondering whether the sequence (-1)^n is converging or diverging. According to Wikipedia, a sequence that doesn't have a limit (i.e, which doesn't converge), is automatically divergent. (-1)^n doesn't have a limit, yet all its values oscillate between 1 and -1, and doesn't tend to ∞. So, is it a divergent sequence or something else?

    Thanks! :D
     
  2. jcsd
  3. Aug 24, 2012 #2
    Does it have a limit?
    No, then it's divergent.
    Divergent means that it doesn't have a number for a limit.
     
  4. Aug 24, 2012 #3
    Divergence is the lack of a limit. If a sequence does not oscillate but instead increases without bound, it is said to be diverging to infinity, or when it is rather obvious, it is said it simply diverges. In the extended real number line, one can say that a sequence converges to infinity.
     
  5. Aug 24, 2012 #4
    Thank you, johnqwertyful and Millenial for your replies, but I'm still in doubt (sorry!).

    So, you mean to say that non-oscillating sequences without a limit are divergent. But what then is an oscillating sequence without a limit? It definitely doesn't converge, and diverges if one takes the definition of divergence as the "absence of a limit". But where does it diverge to?

    For example, consider the sequence (-1)n.n : -1,2,-3, 4, -5, 6,.....

    It seems to increase (as well as decrease) without bound, and doesn't have a limit. So, by the definition of divergence (absence of limit), it does diverge, but towards where? +∞ or -∞? Or does it not matter?

    And can anyone give me a mathematical definition of an oscillating sequence?

    I'm sorry, but this whole conundrum arose from the fact that Thomas-Finney states that (-1)n is divergent, whereas a Maths encylopedia named MATHS 1001, states that it neither converges nor diverges.
     
  6. Aug 24, 2012 #5
    It doesn't have to diverge to anything. If it is explicitly stated that a sequence is divergent, it means it has no limit. If it is loosely stated, and the reader is expected to understand that it grows without bound, it means that the sequence diverges to infinity.

    Oscillating sequences are, by definition, neither convergent nor divergent to infinity. Oscillation as a quantifying measure is defined by the difference of the limit superior and the limit inferior of a function.
     
  7. Aug 24, 2012 #6
    Hi.

    Yes, if it diverges to something, then - it converges to it. Diverges means there is nothing one can point at and say: "Yes, it went that way".

    In math, we use a notion of a [itex]\mathbb{Limit\;Point}[/itex]. Point P is limit point if in EVERY neighbourhood of P there is at list one more point from sequence. Exact techical definition is here: http://en.wikipedia.org/wiki/Limit_point.

    So, [itex](-1)^n[/itex] oscillates from 1 to -1 and back. So Sequence has no points in [itex]\langle -1,1\rangle[/itex]. This segment is part of neighborhoods of 1 and -1 with no other sequence points in them. Obviously, there is no limit point. Same is true if we restrict analysis to natural numbers from [itex]\mathbb{N}[/itex] only.

    Yes, sequences can diverge and stay bounded.

    Cheers.
     
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