# Sequence limit proof

1. Oct 6, 2011

2. Oct 6, 2011

### SammyS

Staff Emeritus
I doubt that he wants an ε - N proof.

Probably wants more than a rough demonstrative "proof". I would.

Do a comparison test. The hint is right there !

3. Oct 6, 2011

### Shackleford

Oh, I know. I just thought that was too easy. I'll work it out shortly.

4. Oct 6, 2011

### Shackleford

xn = (sn)

sn+1/sn = xn+1/xn = x*xn/xn = x

x ≥ L > 1

Eh. I think that "proves" it. Assuming x > 1, then the sequence ratio is bigger than 1. And, of course, the limit is given which is infinity which is what we were asked to show for a sequence that satisfies that ratio.

Last edited: Oct 6, 2011
5. Oct 6, 2011

### SammyS

Staff Emeritus
I It doesn't say xn = (sn) at all.

sn+1 ≥ L sn ≥ L2 sn-1 ≥ L3 sn-2 ≥ ...

What can you say about $\displaystyle \lim_{n\to\infty}L^n\,?$

This leads nicely to an ε - N proof. Maybe do one.

6. Oct 6, 2011

### Shackleford

That limit should go to infinity, right?

The book does the proof for when L < 1. I'm looking over that proof now.

Last edited: Oct 6, 2011
7. Oct 6, 2011

### SammyS

Staff Emeritus
No, it doesn't go to zero. L > 1 !

8. Oct 6, 2011

### Shackleford

Yeah, I didn't mean to put zero.