What is the Limit of a Sequence with L > 1?

[Shackleford]

Does it look like he wants a rough demonstrative "proof" involving plugging in xⁿ = (s_n) or rather, say, an epsilon-N proof?

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png?t=

 

[SammyS]

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 !

 

[Shackleford]

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 !

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

 

[Shackleford]

xⁿ = (s_n)

s_n+1/s_n = 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.

 

[SammyS]

xⁿ = (s_n)

s_n+1/s_n = 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.

I It doesn't say xⁿ = (s_n) at all.

s_n+1 ≥ L s_n ≥ L² s_n-1 ≥ L³ s_n-2 ≥ ...

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

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

 

[Shackleford]

I It doesn't say xⁿ = (s_n) at all.

s_n+1 ≥ L s_n ≥ L² s_n-1 ≥ L³ s_n-2 ≥ ...

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

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

That limit should go to infinity, right?

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

 

[SammyS]

That limit should go to zero, right?

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

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

 

[Shackleford]

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

Yeah, I didn't mean to put zero.