# Unbounded Sequences w.r Divergence

1. Apr 3, 2012

### skunkswks

considering divergence of a sequence in the reals, a$_{n}$, 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:
$\forall$ M $\exists$ N, $\forall$ n$\geq$N $\Rightarrow$ a_n $\geq$ 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. Apr 3, 2012

### DonAntonio

$a_n\rightarrow ∞ \Longleftrightarrow \forall M\in ℝ \,\,\exists N_M\in N\,\, s.t. \,\,n>N_M\Longrightarrow a_n > M$ .

DonAntonio

3. Apr 3, 2012

### skunkswks

yep that is the definition. M>0 could also work fine instead in that defination?

4. Apr 3, 2012

Any M works.

DonAntonio

5. Apr 3, 2012

### Bacle2

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.

6. Apr 4, 2012

### skunkswks

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$_{n}$ → +∞ , does it only have to be Unbounded above (instead of also being increasing)?

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