Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :

    http://tinypic.com/r/11udow4/5

    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] .

    DonAntonio
     
  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.

    DonAntonio
     
  6. Apr 3, 2012 #5

    Bacle2

    User Avatar
    Science Advisor

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Unbounded Sequences w.r Divergence
  1. Unbounded sequence (Replies: 7)

Loading...