Limits at Infinity: Let S(n) Converge to S

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Discussion Overview

The discussion revolves around the behavior of limits at infinity, specifically concerning the convergence of sequences of partial sums in infinite series. Participants explore whether the limit of S(n+1) equals S when the limit of S(n) approaches S, and how this can be formally demonstrated using epsilon-N proofs.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that if the limit of S(n) as n approaches infinity is S, then the limit of S(n+1) must also be S.
  • Others propose that if n is greater than or equal to N, then n+1 is also greater than or equal to N, which implies that both S(n) and S(n+1) are within epsilon of S.
  • A participant mentions the Cauchy property of convergent series, suggesting that the difference S(n+1) - S(n) approaches zero as n approaches infinity.
  • Some participants question whether the limit of S(n-1) also approaches S, proposing a method to verify this using adjusted epsilon values and the triangle inequality.
  • There is discussion about the use of epsilon-N arguments to formalize the reasoning behind these limits, with varying levels of confidence expressed in the application of these proofs.

Areas of Agreement / Disagreement

Participants generally agree that if S(n) converges to S, then S(n+1) also converges to S, but there is some uncertainty about the implications for S(n-1). The discussion remains unresolved regarding the formal verification of these limits and the application of epsilon-N proofs.

Contextual Notes

Limitations include the dependence on the definitions of convergence and the assumptions made about the sequence of partial sums. The discussion also highlights the need for careful handling of indices in the context of limits.

mathsciguy
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I have a question about limits at infinity, particularly, about a limit I have seen in the context of infinite series convergence.

Let's say we have an infinite series where the the sequence of partial sums is given by {S(n)} and also, it is convergent and the sum is equal to S. Then we know that the limit of S(n) as n approaches infinity is S, but from this, can we also say the the limit of S(n+1) is equal to S?

Well, based from the textbook I was reading, it is. I actually have an intuitive idea of why, but I'd rather see something more of a 'formal' explanation, probably something which has epsilons. I tried verifying with an epsilon-N proof but I'm not that confident.
 
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mathsciguy said:
I have a question about limits at infinity, particularly, about a limit I have seen in the context of infinite series convergence.

Let's say we have an infinite series where the the sequence of partial sums is given by {S(n)} and also, it is convergent and the sum is equal to S. Then we know that the limit of S(n) as n approaches infinity is S, but from this, can we also say the the limit of S(n+1) is equal to S?
Yes. Both limits result in S.
mathsciguy said:
Well, based from the textbook I was reading, it is. I actually have an intuitive idea of why, but I'd rather see something more of a 'formal' explanation, probably something which has epsilons. I tried verifying with an epsilon-N proof but I'm not that confident.

It sounds like you've made a good start. People who are new to this try to do things with a ##\delta\ \epsilon## argument. If you can show that for a given ##\epsilon## a number N can be found so that if n ≥ N, then |Sn - S| < ##\epsilon##, you're set. Keep in mind that Sn+1 is the next number in the sequence. As long as the index is past the magic number N, every term in the sequence satisfies the inequality above.
 
Hm, well I could think that obviously if n≥N then this implies n+1≥N (if N>0 and n is restricted to positive integers). Does that mean I can say: if for any ε>0, and N>0 such that |S(n)-S|<ε whenever n≥N is true by hypothesis, then |S(n+1)-S|<ε whenever n+1≥N is also true?

The catch is I've used both N for S(n) and S(n+1).
 
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mathsciguy said:
Hm, well I could think that obviously if n≥N then this implies n+1≥N (if N>0 and n is restricted to positive integers). Does that mean I can say: if for any ε>0, and N>0 such that |S(n)-S|<ε whenever n≥N is true by hypothesis, then |S(n+1)-S|<ε whenever n+1≥N is also true?
It's simpler than that: If n ≥ N, then for sure n+1 ≥ N. So Sn is within ##\epsilon## of S, Sn+1 is within ##\epsilon## of S, Sn+2 is within ##\epsilon## of S, ...
mathsciguy said:
The catch is I've used both N for S(n) and S(n+1).
 
I think that's same things as what I've said. Haha, sometimes I have a hard time wrapping my head around stuff like this. Thanks anyway.
 
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Maybe this summarizes the ideas here:
You want to know :

Limn→oo[S(n+1)-S(n)]

But, since the series converges, the sequence of partial sums is Cauchy, so that
the difference in the parentheses goes to zero, i.e., the limit is 0. If you show quickly that both limits exist (more precisely, you can show that if S(n) exists as n→∞ , then S(n+1) also exists). Then you can say:

Limn→oo[S(n+1)-S(n)]=0 . Since both the limits exist, the limit distributes and

Limn→∞S(n)=Limn→∞S(n+1)
 
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What if the limit of S(n) as n approaches infinity is known to be S, I've read that the function S(n-1) as n approaches infinity is also S. How do we verify this? My idea is that, after finding an N such that: |S(n)-S|<ε whenever n>N; and we adjust ε so that we find a 1<N.

So that: |S(n-1)-S|<ε' whenever (n-1)>(N-1).
 
mathsciguy said:
What if the limit of S(n) as n approaches infinity is known to be S, I've read that the function S(n-1) as n approaches infinity is also S. How do we verify this? My idea is that, after finding an N such that: |S(n)-S|<ε whenever n>N; and we adjust ε so that we find a 1<N.

So that: |S(n-1)-S|<ε' whenever (n-1)>(N-1).

Basically, |S_n- S|<e/2 (S_n converges) , and |S_(n-1)-S_n|<e/2 ( convergence implies

Cauchy) and then the triangle ineq.

kicks-in :

You can use that every convergent sequence is Cauchy, which means that for every e>0 there

is a positive N with |S_n-S_m| <e for all n,m>N . Choose, then, some N > n for any e>0

so that |S_n-S_(n-1)| <e . Then |S_(n-1) -S |=

|S_(n-1) -S_(n)+S_(n)-S | ≤ ...(triangle ineq.)
 
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