Analysis(sequences) proof: multiplying infinite limit at infinity by 0

[K29]

Homework Statement

Let [tex]\stackrel{lim}{_{n \rightarrow \infty}}a_{n} = \infty[/tex]
Let [tex]c \in R[/tex]
Prove that
[tex]\stackrel{lim}{_{n \rightarrow \infty}} ca_{n}=[/tex]

[tex]\infty[/tex] for [tex]c>0[/tex] (i)

[tex]- \infty[/tex] for [tex]c<0[/tex] (ii)

[tex] 0 [/tex] for [tex]c=0[/tex] (iii)

Homework Equations

Definition of divergence to infinity (infinite limit at infinity)
[tex] \forall A \in R. \exists K\in R [/tex] such that [tex]a_{n} \geq A, \forall n \geq K[/tex]

The Attempt at a Solution

For the first two cases I just used the above definition and essentially multiplied c by the inequality.
For the c=0 case used the definition for a finite limit:
[tex]\forall \epsilon > 0 \exists K_{\epsilon} \in R[/tex] such that [tex]\forall n \in N, n \geq K_{\epsilon}, |a_{n}-L|<\epsilon[/tex]
Now if I can squeeze [tex]0 \leq |c a_{n}-0| \leq ?=0[/tex] then I'm done
But I can't see an upper limit for the inequality.Stuck there.
Or is there a way to prove this by contradiction instead of the way I've chosen.
Help?

 

[LCKurtz]

But in the case where c = 0, what is ca_n?

 

[K29]

Its 0.So sure, the sandwich theorem can't work. So could it be as simple as using the definition of divergence to get contradicion.
I must have [tex]ca_n \geq A, \forall A[/tex] but by fixing n I get [tex]ca_{n}=0 < A[/tex]. Contradiciton.
But is it enough to say that it does not diverge to infinity and minus infinity therefore it must converge to zero? Surely not?

 

[K29]

Oh wait I think I see somthing else
Using the theorem that says
If [tex]|a_{n}-L|=0 \forall n[/tex] then [tex]|a_{n}-L| < \epsilon[/tex] etc etc defn of limit.
So I could prove by induction that |[tex]ca_{n}-0|=0[/tex] or more simply that
[tex]ca_{n}=0 \forall n[/tex] But then I get stuck on what to do with
[tex]ca_{n+1}[/tex]

 

[LCKurtz]

But in the case where c = 0, what is ca_n?

Its 0.So sure, the sandwich theorem can't work.

You are trying to make |ca_n - 0| < ε

If ca_n = 0, how hard is that?

 

[K29]

I thought I'd have to prove [tex]ca_{n} = 0 \forall n[/tex] by induction. But I've thought about it and I think I see that I shouldn't need to
Thanks