Convergence of a sequence when ratio of consecutive term converges

[vrbke1007kkr]

Homework Statement

Assume all s_n =/= 0, and that the limit L = lim|(s_n+1)/(s_n)| exist.

a) Show that if L<1 then lim s_n = 0

b) Show that if L>1 then lim s_n = + oo

Homework Equations

There is a hint that we should select 'a' s.t. L<a<1 and obtain N s.t. n>N => |s_n+1 < a|s_n|.

Then show that |s_n| < a^n-N|s_N|

The Attempt at a Solution

Its easy to get: |s_n+1|< ($$\epsilon+L$$)|s_n|, now I tried using the fact that L<1 to find an epsilon and an 'a' such that |s_n+1|< a|s_n|
Even if I find such a, and even if I can prove |s_n| < a^n-N|s_N|, how do I get that |s_n| < $$\epsilon$$ for all epsilon

 

[Dick]

Homework Statement

Assume all s_n =/= 0, and that the limit L = lim|(s_n+1)/(s_n)| exist.

a) Show that if L<1 then lim s_n = 0

b) Show that if L>1 then lim s_n = + oo

Homework Equations

There is a hint that we should select 'a' s.t. L<a<1 and obtain N s.t. n>N => |s_n+1 < a|s_n|.

Then show that |s_n| < a^n-N|s_N|

The Attempt at a Solution

Its easy to get: |s_n+1|< ($$\epsilon+L$$)|s_n|, now I tried using the fact that L<1 to find an epsilon and an 'a' such that |s_n+1|< a|s_n|
Even if I find such a, and even if I can prove |s_n| < a^n-N|s_N|, how do I get that |s_n| < $$\epsilon$$ for all epsilon

If you have |s_n|<a^(n-N)*|s_N|, what do you get if you take the log of both sides? What does this tell you about log(|s_n|) as n->infinity?

 

[vrbke1007kkr]

I get log(s_n) < (n-N)log(a) + log(S_N), since n>N as n-> oo, the RHS goes to infinity?! I was hoping that the RHS would go to -oo so that we may conclude |S_n| -> 0

 

[Dick]

I get log(s_n) < (n-N)log(a) + log(S_N), since n>N as n-> oo, the RHS goes to infinity?! I was hoping that the RHS would go to -oo so that we may conclude |S_n| -> 0

L<a<1, right? What sign is log(a)?

 

[vrbke1007kkr]

L<a<1, right? What sign is log(a)?

a negative number... wow I really need to review properties of these functions... thanks

Now the only thing is, how do I find such an 'a'? What epsilon can I choose so that $$\epsilon$$ + L < 1, when L<1

It seems like any linear combination of L is out of the question, powers of L doesn't seem to fit neither.

 

[Dick]

a negative number... wow I really need to review properties of these functions... thanks

Now the only thing is, how do I find such an 'a'? What epsilon can I choose so that $$\epsilon$$ + L < 1, when L<1

It seems like any linear combination of L is out of the question, powers of L doesn't seem to fit neither.

You don't need to find any epsilon. If L<1, then there really isn't any problem to find an 'a' such that L<a<1. What's wrong with (L+1)/2? I think you are overthinking this.

 

[vrbke1007kkr]

You don't need to find any epsilon. If L<1, then there really isn't any problem to find an 'a' such that L<a<1. What's wrong with (L+1)/2? I think you are overthinking this.

Finding an L<a<1 isn't difficult, but don't I have to find an N such that n>N implies |S_n+1| < a|s_n|?

From the definition of a limit I got: there exist N s.t. n>N => ||s_n+1/s_n| - L|< e

which means |s_n+1|<(e+L)*|s_n|,

I guess we don't need to assign e (epsilon) to be any particular value to see that we can definitely make e small enough so that e+L < a = (L+1)/2... (not sure if the prof would buy such an argument though)

I get the feeling that I'm missing a huge point here

 

[vrbke1007kkr]

On a second thought, I think I kind of get it.

If we let e (epsilon) < 1 - L, and a = e + L, e>0, then obviously L<a<1

Please let me know if there is anything wrong with the following thoughts:

Since |s_n+1|<a|s_n| for all n>=N, then |s_N+1|<a|s_N|, and |s_N+2|<a|s_N+1|=a|a|s_N||, so on and so forth. Then its a matter of algebra to show that |s_n|<a^(n-N)|s_N|, and then we are pretty much done?

 

[Dick]

On a second thought, I think I kind of get it.

If we let e (epsilon) < 1 - L, and a = e + L, e>0, then obviously L<a<1

Please let me know if there is anything wrong with the following thoughts:

Since |s_n+1|<a|s_n| for all n>=N, then |s_N+1|<a|s_N|, and |s_N+2|<a|s_N+1|=a|a|s_N||, so on and so forth. Then its a matter of algebra to show that |s_n|<a^(n-N)|s_N|, and then we are pretty much done?

That sounds ok to me.