Limit of a sequence

  • Thread starter Kamataat
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  • #1
Kamataat
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Hi!

Have I understood the concept of the limit of a sequence correctly?

Let's say we have a sequence {x_n} (dots for spacing):

----|------------------|-------------------|----------|------->
a-epsilon......a.....a+epsilon...R...x_n

So "a" is the limit of {x_n} if there exists an n_0 = f(epsilon) so that for every n > n_0 the inequality |x_n - a| < epsilon is true.

How can |x_n - a| < epsilon be true for values of n which go beyond a+epsilon? For example, for R the inequality |R - a| < epsilon is not true, yet the definition of a limit of a sequence says that the inequality must hold for all n > n_0 (including that n which gives us R).

I guess they avoid that problem on the "left side" of "a" by saying that n_0 is a function of epsilon and that automatically cuts off all values of "n" before a-epsilon. Why is that not done on the "right side"? They allow "n" to increase towards positive infinity so that the inequality won't hold past the value of "n" which gives us a+epsilon.

- Kamataat
 

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  • #2
arildno
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No, no!
You must have:
[tex]|x_{R}-a|<\epsilon[/tex]
[tex]x_{R}[/tex] represents the R'th term in the sequence, and there is no reason why this term should equal R.
 
  • #3
Kamataat
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Yes, I know.

In my inequality |x_n - a| < epslion the "x_n" term means n-th element of the sequence.

Why can't it equal R? The definition says that for all n > n_0. If n_0-th element of the sequence gives us the value a-epsilon, then if we start increasing n, we can get values between a-epsilon and a, a and a+epsilon and values greater than a+epsilon. But for values greater than epsilon, the inequality is no longer true. But "a" to be the limit, the inequality has to be true for all n > n_0 including those which give us values greater than a+epsilon. But it can't be true for values greater than a+epsilon.

- Kamataat
 
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  • #4
matt grime
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But, there need not be any particular n such that x_n=R, indeed by definition, if x_n is convergent etc.. there is no n greater than n_0 such that x_n=R.

I don't think you understand what a sequence is from the looks of it.


A sequence is a set of elements of, in this case, the real numbers indexed by the natural numbers, eg

x_n = 1/n,

or
x_n = 2 for all n


Both of those converge, the first to zero the second to 2.



It doesn't make sense to say "those values of n which go beyond a+e", at least not to me: what goes beyond a+e? x_n? why must it?
 
  • #5
Kamataat
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matt grime said:
But, there need not be any particular n such that x_n=R, indeed by definition, if x_n is convergent etc.. there is no n greater than n_0 such that x_n=R.

I don't think you understand what a sequence is from the looks of it.

It doesn't make sense to say "those values of n which go beyond a+e", at least not to me: what goes beyond a+e? x_n? why must it?

I do understand what a sequence is.

When I say "those values of n which go beyond a+e" I mean values of n which give us elements of the sequence with values greater than a+e.

Grrrrrrrrr. My bad. I understand it now (I hope). For some reason I was thinking that a can be in the middle of the sequence. GRRRRRRRRRRRRRRRRRRRRRRRRRRRR!

So if the sequence converges to a then there is no n > n_0 that gives us values of the sequence's elements greater than a (or on the right side of a)?

- Kamataat
 
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  • #6
matt grime
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No, if x_n converges to a, and we've chosen n_0 given that e, then there is no x_n>a+e for n>n_0

1/n converges to zero, yet all terms in the sequence are greater than 0.
 
  • #7
shmoe
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Kamataat said:
I
So if the sequence converges to a then there is no n > n_0 that gives us values of the sequence's elements greater than a (or on the right side of a)?

That's not correct. There is no n>n_0 where x_n is greater than a+epsilon, x_n could still be greater than a.
 
  • #8
Kamataat
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How can x_n be greater than a? Doesn't the fact that a is the limit of x_n mean that x_n converges to a?
 
  • #9
matt grime
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You agree that 1/n converges to zero? And what's special about greater than? why not less than too, since it is just a mirror image, and thus you're effectively saying that only constant sequences converge. Doesn't the fact that your objection has no mention of epsilon worry you?
 
  • #10
Kamataat
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Yes, I agree.

I was talking about greater than beacuse I was assuming an increasing sequence.

Maybe I should ask differently.

If we approach the limit from one side, then can there be any values of the sequence on the other side of that limit? So for an increasing sequence for which a=5, can there be values of x_n > 5? As I understand it, the answer is no.

- Kamataat
 
  • #11
matt grime
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Why didn't you say the sequence was increasing?

If x_n is a strictly increasing sequence then obviously the limit must be greater than any of the elements in the sequence, if the limit exists.
 
  • #12
Kamataat
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I thought you guys would get it out of the context, but I'm sorry if I wasn't clear enough.

At last I get it. It seems obvious to me too.

However, now I have another question. Why e>0? For an increasing sequence, that starts with negative values, passes 0 and converges to a positive value, couldn't I choose a negative e? I mean, does x_n0 = e?

- Kamataat
 
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  • #13
matt grime
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Because the correct statement is

|x_n-a|<e

right?

Now abs values are always positive, so if you picked e negative, then it would make no sense.
 
  • #14
Kamataat
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I edited my post, but in case you didn't notice (I edited it shortly before you replied) I'll ask again:

Does x_n0 = e or does x_n0 = a+/-e (depending on whether the sequence is increasing or decreasing)?

- Kamataat
 
  • #15
matt grime
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What is x_n0? My post still is the answer to why we have e positive: limits use abs values, if e were negative in that definition, we would asking for something that must be positive to be strictly negative, and that is contradictory. Again, look back at where the e, for epsilon, arises.
 
  • #16
Kamataat
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x_n0 is the n_0-th term of the sequence.

So basically I'm asking that if I draw the sequence on an axis then where on that axis would e be?

- Kamataat
 
  • #17
matt grime
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How do you know x_n0 =e? At what point do you see equality in any of the parts of the definition of limit? e does not "live in the same space" as the elements in the sequence.

x_n tends to a if for all e>0, there is an n_0 such that |x_n-a| < e for all n> n_0

at no point am I requiring x_n_0 to equal e, or to equal a+/-e, I'm not even declaring that a+e even makes sense, only that there is a distance notion in the space where we're talking about taking limits.

It is only in the special case of Real numbers that we can talk about a+/-e.

Where e is on the real axis is entirely immaterial to this argument.


So, 1/n tends to zero, obviousl, so Iknow that for e=1/sqrt(2) etc.. even though there is no element in the sequence with n_0= 1/sqrt(2).
 
  • #18
Kamataat
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1.) Well, since I am currently only dealing with real numbers, then can I assume a+/-e?

2.)
x_n tends to a if for all e>0, there is an n_0 such that |x_n-a| < e for all n> n_0

If e can be outside of the "space of the sequence" then what need is there for n_0 to be dependent on e? Couldn't I just as well say that "x_n tends to a if there is an n_0 such that |x_n-a| < e for all n> n_0"?
- Kamataat
 
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  • #19
matt grime
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You can assume a+/e is what, does what, erm what?
 
  • #20
matt grime
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You could say that. However almost always the n_0 will be dependent on e.

Have you considered thinking about what's going on with some simple examples?
 
  • #21
Kamataat
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That the n_0-th element of {x_n} equals a+/-e.

- Kamataat
 
  • #22
matt grime
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NO! there is absolutely no reason why any of the elements in the sequence is equal to a+/e. epsilon is absoutely arbitrary (apart from bein positive). That is the entire point of limits.
 
  • #23
Kamataat
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Ok, let's take an increasing sequence that converges to a. Below the axis are the values of the elements above the axis.

.....x_n0......x_n as n --> +infinity
----------|--------------------|
.....a-e.....a

Would that be right when dealing with real numbers only?

edit: upon seeing your last post, i guess that ascii graphic is wrong. or could i assume the n_0-th to be equal to a-e in the case of real numbers even though it wouldn't have to be in a broader sense?

- Kamataat
 
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  • #24
matt grime
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And so?

Look at the definition again. Where in the definitio does it even mention that x_n0-a=e? It doesn't given ANY e, x_n converges to a iff there is an n_0 such that

|x_n-a|< e for all n>n_0

I don't ever ask that given any e I can find n_0 such that x_n_0=a+/-e. I don't need to ever ask that that HAS to happen.

For simplicity, I'm going to keep using the sodding concrete example that you keep ignoring. 1/n tends to zero, right?

Why? (And no one point out the egregious error in this, please, it's not the point in hand)


Because, give me any epsilon, and I have to mean ANY e>0, consider 1/e, take the ceiling of this, that is the next largest integer, let that be n_0.

For all n>n_0>1/e it follows that |1/n - 0| <e

e was arbitary,hence it converges to zero.

There is no point that I claim, or need to claim that 1/n_0 = e.

Do you follow that example?

Anyway, I'm off to the pub.
 
  • #25
Kamataat
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I follow that.

- Kamataat
 
  • #26
Kamataat
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Despite that I was able to follow that last example using many random values for e, I still do not understand why we have to make n_0 dependent on e. Since e a completely random positive number that gives us n_0, then why not do away with e and just pick n_0 > 0 randomly?

- Kamataat
 
  • #27
Kamataat
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Ok, e is COMPLETELY arbitary. But can the dependence of n_0 on e, n_0(e), also be arbitarily chosen (as long as we make sure that n_0 is positive)?

- Kamataat
 
  • #28
matt grime
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I really don't get those questions in this sense.

the e is very necessary in this definition of convergence, you cannot do away with e.

Also you do not "choose e" as your last post implicitly states.

n_0 may not be picked to be any number, there is a condition it must satisfy, that all terms after the n_0'th are no more than e away in distance. So it clearly will depend on e. If you did understand the example, where I pick n_0 to be ceiling(1/e) then you'd have surely not asked that question.

I mean suppose e were 1/10,000 then I can't pick n_0 arbitrarily and independently of e can I? say I set n_0 =1, then it is not true that for all n>1 that |1/n - 0| < 1/10,000

eg pick n any number from 2 to 10,0000.


The form of convergence proofs is: given any e>0, we must find an n_0 such that...

quantify it: for all e>0 there exists n_0 in N such that...

what would you suggest doing instead?
 
  • #29
Kamataat
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After thikig about it for a day, I guess I'll try some examples from my textbook. It seems clearer now. Anyway, thanks for all your time! I won't know whether I understand limits until I've tried some excercise problems, but I know your help made it a lot clearer to me!

- Kamataat
 

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