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Unbounded Sequences w.r Divergence

  1. Apr 3, 2012 #1
    considering divergence of a sequence in the reals, a[itex]_{n}[/itex], if such a sequence → +∞ as → n, then I would like to know what type of sequence this reuqires. (excluding divergence to -∞ for now)

    so a_n → +∞ iif:
    [itex]\forall[/itex] M [itex]\exists[/itex] N, [itex]\forall[/itex] n[itex]\geq[/itex]N [itex]\Rightarrow[/itex] a_n [itex]\geq[/itex] M .

    So is the above equivalent to stating ( and so another way of saying a_n → +∞ ):
    1. a_{n} is increasing &
    2. a_{n} NOT bounded above ?

    now my main question is, why can't i simply say a_n → +∞ iff a_{n} is NOT BOUNDED ABOVE (and nothing else).

    surely then a_{n} by the definition of being unbounded above means a_{n}has no choice but to increase towards +∞? Right...?

    and one more consideration: so then a_{n} could be something like :


    as drawn. This sequence oscillates, diverges and heads to + ∞ as well as -∞. So can I say this sequence → ∞ or -∞ or which!?

    Thanks for any help.
    Last edited: Apr 3, 2012
  2. jcsd
  3. Apr 3, 2012 #2
    [itex]a_n\rightarrow ∞ \Longleftrightarrow \forall M\in ℝ \,\,\exists N_M\in N\,\, s.t. \,\,n>N_M\Longrightarrow a_n > M[/itex] .

  4. Apr 3, 2012 #3
    yep that is the definition. M>0 could also work fine instead in that defination?
  5. Apr 3, 2012 #4

    Any M works.

  6. Apr 3, 2012 #5


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    I don't understand, aren't you using M as a variable, not a constant?

    Re your function, if the pattern extends to infinity, then the function does not have a limit of oo.
  7. Apr 4, 2012 #6
    Okay M as a variable then, but im just trying to specfiy what range of values it can take.

    So with my original post, for a[itex]_{n}[/itex] → +∞ , does it only have to be Unbounded above (instead of also being increasing)?

    and with the defination of a[itex]_{n}[/itex] → +∞, that function would not 'diverge to +∞' right?
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