- #1
skunkswks
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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.
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.
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