Sequence inequality, epsilon N argument

[flyingpig]

Homework Statement

I already have the solutions, but I am not sure what the solutions are trying to say.

[PLAIN]http://img194.imageshack.us/img194/2595/unledlvc.jpg

So in

If N is any integer greater than \frac{1}{\epsilon}, the implication n > N will hold

I don't understand this, we have

n > \frac{1}{\epsilon}

and If (and I am guessing we really want this to hold)

N > \frac{1}{\epsilon}

How does that deduce n > N? Wouldn't it make more sense to say that N < \frac{1}{\epsilon} and hence N < \frac{1}{\epsilon} < n?

One final question from me is, how does N relate to epsilon? I understand that

|a_n - L | < \epsilon is speaking how close a_n is to our limit L for a very very small epsilon and the goal of the proof-theorem is to find such an epsilon, what role does n > N play with respect to |a_n - L | < \epsilon

I have accompanied this with a picture

[PLAIN]http://img213.imageshack.us/img213/5001/unledsy.jpg

I cna see that the epsilon is trying to "squeeze" our limit and how n > N is supporting this too. So is the intuition that they (both inequalities) are saying the same thing but they are both supporting each other?

I'll clarify if possible

 

[Mark44]

Suppose ε is given as 1/10. Then we can take N = 10. Then for any integer n for which n > N, 1/n < ε. That's really all the above is trying to say.

For example:
n = 11, 1/11 < 1/10
n = 12, 1/12 < 1/10
and so on.

In answer to your question, if we take N < 1/ε, as you suggest, then using my example value of ε = 1/10, we could take N = 8 (or N = 5 are whatever, as long as N < 10). That will cause problems, because 1/9 > 1/10 = ε.

 

[flyingpig]

Suppose ε is given as 1/10. Then we can take N = 10. Then for any integer n for which n > N, 1/n < ε. That's really all the above is trying to say.

Now I am even more confused because the conditions were N &gt; \frac{1}{\epsilon} or N &lt; \frac{1}{\epsilon}

Take ε = 1/10, in either case you get something like this

N = 10 &gt; \frac{1}{\frac{1}{10}} = 10, i.e. 10 > 10 and WLOG, 10 < 10, which is absurd in either case (I don't know why my book uses "absurd" when we do contradictions proofs, it seemed kinda rude)

For example:
n = 11, 1/11 < 1/10
n = 12, 1/12 < 1/10
and so on.

Do you actually mean ε = 10 instead?

In answer to your question, if we take N < 1/ε, as you suggest, then using my example value of ε = 1/10, we could take N = 8 (or N = 5 are whatever, as long as N < 10). That will cause problems, because 1/9 > 1/10 = ε.

So I thinky ou really do mean ε = 10 and where did you get 1/9 from this time?

thanks

 

[Mark44]

No, I don't mean ε = 10. ε is generally a "small" number, typically much smaller than 1. The smaller ε is, the farther out in the sequence you have to go to find terms in the sequence that are smaller than ε.

You want N > 1/ε. (Some texts will say N >= 1/ε.)

If ε = .01, then 1/ε = 100, and we can take N = 100.

Now for any n > N, 1/n < ε.

You're really overthinking this.

Where are you getting that the conditions include N < 1/ε? There's nothing in your OP that says this.

 

[flyingpig]

In answer to your question, if we take N < 1/ε, as you suggest, then using my example value of ε = 1/10, we could take N = 8 (or N = 5 are whatever, as long as N < 10). That will cause problems, because 1/9 > 1/10 = ε.

How did you arrive at N < 10 again?

edit: you got it from n > N right? So n > 10 and n > N, then N <= 10 which is consistent with my initial condition (not from book)

Where are you getting that the conditions include N < 1/ε? There's nothing in your OP that says this.

[strike]No I made that condition up because I thought should work, but I don't follow your counterexample from above at all. [/strike]
One of the conditions we needed was n > N, so that 1/n < 1/N, can assume there exists this N that n > N? Because if we don't, then it wouldn't work in either case (with more deduction)

When I suggested that N < 1/ε, then 1/N > ε <=> ε < 1/N

So we have ε < 1/N and ε > 1/n => 1/n < 1/N => n > N, so now we could confirm with the assumption n > N.

Do you see where I am going with this?

 

[Mark44]

How did you arrive at N < 10 again?

I didn't. I was trying to show you why what you were doing was incorrect. This is what I said (emphasis added).

In answer to your question, if we take N < 1/ε, as you suggest, then using my example value of ε = 1/10, we could take N = 8 (or N = 5 are whatever, as long as N < 10). That will cause problems, because 1/9 > 1/10 = ε.

edit: you got it from n > N right? So n > 10 and n > N, then N <= 10 which is consistent with my initial condition (not from book)



[strike]No I made that condition up because I thought should work, but I don't follow your counterexample from above at all. [/strike]



One of the conditions we needed was n > N, so that 1/n < 1/N, can assume there exists this N that n > N? Because if we don't, then it wouldn't work in either case (with more deduction)

The way this limit process works is as a dialogue. You're trying to prove that \lim_{n \to \infty} \frac{1}{n} = 0.

Let's say I'm skeptical, and don't believe you.

I say, OK, can you find some point in the sequence beyond which all of the following numbers are < 1/100 (that's ε).

You do a quick calculation, and say sure. You're thinking, I want to find a number N that works. So, because of this particular sequence, you calculate N = 1/ε = 100.

And then you show me that all of the terms from a₁₀₁ on are smaller than ε.

If I'm still not satisfied, I'll come back, "Can you make them smaller than 1/1000?" Using the same process as before, you tell me that all of the terms from a₁₀₀₁ on are smaller than ε.

The idea is that you want to find an index N so that on one side, all of the terms are smaller than ε. That means all of the terms whose indexes n are larger than N. On the other side of n, the terms in the sequence are >= ε, so those terms don't serve to prove that the sequence is converging to whatever the limit is supposed to be.

When I suggested that N < 1/ε, then 1/N > ε <=> ε < 1/N

So we have ε < 1/N and ε > 1/n => 1/n < 1/N => n > N, so now we could confirm with the assumption n > N.

Do you see where I am going with this?