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

[mathsciguy]

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.

 

[Mark44]

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.

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 |S_n - S| < $\epsilon$, you're set. Keep in mind that S_n+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.

 

[mathsciguy]

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).

 

[Mark44]

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 S_n is within $\epsilon$ of S, S_n+1 is within $\epsilon$ of S, S_n+2 is within $\epsilon$ of S, ...

The catch is I've used both N for S(n) and S(n+1).

 

[mathsciguy]

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.

 

[Bacle2]

Maybe this summarizes the ideas here:
You want to know :

Lim_n→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:

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

Lim_n→∞S(n)=Lim_n→∞S(n+1)

 

[mathsciguy]

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).

 

[Bacle2]

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.)