# Convergence of two limits (Analysis)

1. Jan 20, 2008

### MrGandalf

Hello. I'm using T.W. Korners 'A Companion to analysis', and I'm struggling with the exercises. Never been interested in proofs or how to derive them, so I guess I'm in for a tough semester.

1. The problem statement, all variables and given/known data
Prove that the first few terms of a sequence do not affect convergence.
Formally, show that if there exists an N such that $$a_n = b_n$$ for $$n \geq N$$, then $$a_n \rightarrow a$$ as $$n \rightarrow \infty$$ implies $$a_n \rightarrow b$$ as $$n \rightarrow \infty$$.

2. Relevant equations
In the text we just prooved the uniqueness of the limit.
(i) If $$a_n \rightarrow a$$ and $$a_n \rightarrow b$$ as $$n \rightarrow \infty$$, then $$a = b$$.

3. The attempt at a solution
Since we have $$a_n = b_n$$ for $$n\geq N$$ we can use (i) to prove that the limit is the same since the sequences coincide.

Can someone with a bigger brain than mine confirm that this is correct? If not, could you please point out where my reasoning fails?

Thanks!

Last edited: Jan 20, 2008
2. Jan 20, 2008

### morphism

It looks to me like you just reworded the problem. I think the point here is to actually use the formal definition of convergence.

3. Jan 20, 2008

### HallsofIvy

Staff Emeritus
Don't know about a "bigger brain". More experience perhaps.

Your theorem only says that if a sequence converges it can't have two different limits. You can't define {bn} to be a different sequence and then declare that it is the same sequence! The two sequence do not "coincide" until after "N" and these two different sequences have the same limit is what you want to prove!

The definition of convergence is: {an} converges to L if and only if, for all $\epsilon> 0$ there exist M (I'm using M here because you are already using N for a different purpose) such that if n> M then |an- L|< $\epsilon$. Now, how big do you think M should be relative to your N?

Last edited: Jan 20, 2008
4. Jan 20, 2008

### MrGandalf

Thank you, I'll get back to this exercise later and see what I can do.

Btw. I probably have the bigger brain. My head is exceptionally large. :tongue:

5. Jan 20, 2008

### MrGandalf

After further straining my enormous noggin, I noticed another lemma, which I think can be applied to this exercise.

(ii) If $$a_n \rightarrow a$$ as $$n \rightarrow \infty$$ and $$1 \leq n(1) < n(2) < n(3) < \dots$$, then $$a_{n(j)} \rightarrow a$$ as $$j \rightarrow \infty$$

This one basically says that if we have a sequence tending to a limit, we can take a subset of that sequence, and make a new subsequence. The subsequence will converge to the same limit as the supersequence.

Returning to the exercise: I have two sequences $$a_n$$ and $$b_n$$ that I know are equal when $$n \geq N$$. I make two new subsequences, that in fact will be the same because we choose them to be (we just start the subsequences from N). From (ii) we know that they will have the same limits as the supersequence, and from (i) we know that they will have the same limit, because this time they are the same sequences and the limit is unique.

Just wanted to try my way a little more.

Last edited: Jan 20, 2008
6. Jan 20, 2008

### morphism

That certainly works. But still, at this level I think it's important to prove this directly from the definition, if only for the experience.

7. Jan 20, 2008

### arildno

You might consider a set up like this:
$$|a_{n}-b|=|(a_{n}-b_{n})+(b_{n}-b)|\leq{|}a_{n}-b_{n}|+|b_{n}-b|$$

8. Jan 20, 2008

### MrGandalf

Hmm, I'll have a go.

Well, for $$n\geq N$$, we know that $$a_n = b_n$$, so $$a_n - b_n = 0$$

$$|a_{n}-b|=|(a_{n}-b_{n})+(b_{n}-b)| = |0 +b_{n}-b| = |b_n - b|$$

$$|a_{n} - b| = |b_{n} - b|$$

In the same way we can also deduce
$$|a_{n} - a| = |b_{n} - a|$$

They will fulfill the exact same requirements for convergence, so the limits must be equal.

Last edited: Jan 20, 2008
9. Jan 20, 2008

### HallsofIvy

Staff Emeritus
It's really very simple- just a slight change in how you choose "N" for a given $\epsilon$.

By the way, be sure that you prove this both ways:\

Suppose that, for some N, {an} and {bn} are identical for all n> N. Then

1) If {an} converges then so does {bn}

2. If {an} does not converge then {bn} n does not converge.
Fortunately, that is identical to "if {bn} converges then so does {an}.

10. Jan 21, 2008

### MrGandalf

We have two sequences $$\{a_n\}$$ and $$\{b_n\}$$ such that
(*) $$a_n = b_n$$ for all $$n \geq N_1$$.

From the definition of limits, we know that
$$a_n \rightarrow a$$ when $$n \rightarrow \infty$$ for some $$\epsilon$$ when $$n \geq N_2$$

As I did in my previous post, using (*):
$$|a_{n}-a|=|(a_{n}-b_{n})+(b_{n}-a)| = |0 +b_{n}-a| = |b_n - a|$$
$$|a_{n}-a| = |b_{n}-a|$$ for the same $$n$$.

and conversely
$$|b_{n}-a|=|(b_{n}-a_{n})+(a_{n}-a)| = |0 +a_{n}-a| = |a_n - b|$$
$$|b_{n}-a|=|a_{n}-a|$$ for the same $$n$$

They both fulfill the same requirements for the definition, and therefore they must have the same limit.

I did it right here, didn't I?

Last edited: Jan 21, 2008
11. Jan 21, 2008

### HallsofIvy

Staff Emeritus
I don't see any need for the "triangle inequality" here.

And since it has been some time since this was originally posted, what's wrong with this:

We are given that {an} converges to b. That means that, given any $\epsilon> 0$ there exist some M such that if n> M then $|a_n- b|< \epsilon$.

We are also given that if n> N, then an= bn. Now what can you say about |bn- b| if n> max(N, M)?

12. Jan 21, 2008

### MrGandalf

Since $a_n = b_n$, it follows that $|b_n - b| = |a_n - b|$, so $|b_n - b| < \epsilon$ and we have shown that the sequence bn converges to the same limit as an.

Please confirm if I'm doing it right. May I also ask if my previous post was wrong, or just clumsy? I find this a bit difficult, and I'm not really able to determine these things by my self yet.