Convergence of a sequence when ratio of consecutive term converges

In summary: And you already have that algebraic proof in your posts above. Just be careful with notation- when you use |s_n|, you mean |s_N|, right?Yes, I meant |s_N|. Thanks. I'm done then.
  • #1
vrbke1007kkr
10
0

Homework Statement


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

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

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

Homework Equations


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

Then show that |sn| < an-N|sN|

The Attempt at a Solution


Its easy to get: |sn+1|< ([tex]\epsilon+L[/tex])|sn|, now I tried using the fact that L<1 to find an epsilon and an 'a' such that |sn+1|< a|sn|
Even if I find such a, and even if I can prove |sn| < an-N|sN|, how do I get that |sn| < [tex]\epsilon[/tex] for all epsilon
 
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  • #2
vrbke1007kkr said:

Homework Statement


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

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

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

Homework Equations


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

Then show that |sn| < an-N|sN|


The Attempt at a Solution


Its easy to get: |sn+1|< ([tex]\epsilon+L[/tex])|sn|, now I tried using the fact that L<1 to find an epsilon and an 'a' such that |sn+1|< a|sn|
Even if I find such a, and even if I can prove |sn| < an-N|sN|, how do I get that |sn| < [tex]\epsilon[/tex] 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?
 
  • #3
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
 
  • #4
vrbke1007kkr said:
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)?
 
  • #5
Dick said:
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 [tex]\epsilon[/tex] + 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.
 
  • #6
vrbke1007kkr said:
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 [tex]\epsilon[/tex] + 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.
 
  • #7
Dick said:
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 => ||sn+1/sn| - L|< e

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

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
 
  • #8
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?
 
  • #9
vrbke1007kkr said:
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.
 

Related to Convergence of a sequence when ratio of consecutive term converges

1. What is a sequence?

A sequence is a list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term.

2. What does it mean for a sequence to converge?

A sequence converges when its terms approach a specific value as the sequence continues. This value is called the limit.

3. What is the ratio of consecutive terms in a sequence?

The ratio of consecutive terms in a sequence is the quotient when dividing one term by the term that comes after it. It can also be written as a fraction, with the denominator being the term and the numerator being the term that follows it.

4. How do you determine if the ratio of consecutive terms converges?

If the limit of the ratio of consecutive terms exists and is a finite number, then the ratio converges. This means that as the sequence continues, the terms get closer and closer to this specific number.

5. What is the significance of the ratio of consecutive terms converging in a sequence?

When the ratio of consecutive terms converges, it means that the terms in the sequence are getting closer and closer to each other. This can help determine the behavior and limit of the sequence as a whole.

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