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So I can state the definition:

A sequence of real numbers is said to converge to a Real number $a$ iif for every $\epsilon \gt 0$ there is an $N \in \mathbb{N}$ such that.

$n\geq\mathbb{N}$ implies $\vert x_n - a \vert \lt \epsilon$

I don't understand what $n \geq \mathbb{N}$ means and it's connection to $\vert x_n - a \vert$.

Given a sequence I am supposed to show that the limit exists by using the archimedian principle which I'm not sure how to do with this.

For example:

$x_n = 2-\frac{1}{n}$

$x_n \to 2$ as $n \to \infty$

$\vert 2 - \frac{1}{n} - 2 \vert \lt \epsilon$

$\vert -\frac{1}{n} \vert \lt \epsilon$

Which implies $\frac{1}{n} \lt \epsilon$?

$\frac{1}{\epsilon} \lt n$

I have no idea if this is even remotely close. Can someone explain in simple terms what it is I am exactly supposed to do and why. I am bad with technical definitions until I have some foundation.

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    $\begingroup$ Where did you find that definition? $\endgroup$ Commented Jul 27, 2019 at 19:02
  • $\begingroup$ There is no meaning to $n\geq\mathbb{N}$. It should be: there is $N\in\mathbb{N}$ such that $n\geq N\implies |x_n-a|<\epsilon$. $\endgroup$
    – Mark
    Commented Jul 27, 2019 at 19:03
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    $\begingroup$ You've got the definition wrong. It probably says there is an $N\in\mathbb{N}$ such that $n\geq N\implies|x_n-a|<\varepsilon $ $\endgroup$
    – saulspatz
    Commented Jul 27, 2019 at 19:04
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    $\begingroup$ wouldn't that be any $N \gt \frac{1}{\epsilon}$? $\endgroup$
    – K. Gibson
    Commented Jul 27, 2019 at 19:12
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    $\begingroup$ You have already found the $N$, in the sense you have shown that it exists and that's what you were asked to do. You can compute it concretely for concretely given $\varepsilon>0$ $\endgroup$
    – B.Swan
    Commented Jul 27, 2019 at 19:17

2 Answers 2

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I am going to solve your example and try to explain in each step what is going on. First I want to try to explain what the definition is saying, since it took me about three months myself to understand the epsilon-delta type of definitions. So here goes.

Lets say we have a sequence $\{ a_n\}_{n=1}^{\infty}$. In simple-text the definition of convergence of a sequence means this. A sequence is said to converge to some number $\mathbf{a}$, then from some point somewhere really far down ( far down meaning for some really really large sequence index $n$ ) in the sequence and onwards it is very close to this number.

Now there are few things in this definition which really need to be formalized so that we can have a proper definition of convergence. The first thing is "very close" part of the definition. The very close part is formalized by saying that for arbitrary number $\varepsilon>0$ there is some element in the sequence, call this element $a_k$ such that distance between this element and the number we are converging to is less than $\varepsilon$. Basically said this "very close" means for every $\varepsilon > 0$ we can find some $k$ such that $|a_k - \mathbf{a}| < \varepsilon$.

Next thing we need formalized is "somewhere far down in the sequence". This is formalized by saying that for some large natural number $N \in \mathbb{N}$ we are looking at sequence indexes $n$ such that $n>N$. Its that simple.

Last part we need to formalize is from somewhere in the sequence and onwards. This means that after we have picked our large point $N\in \mathbb{N}$ that sequence must be close to $\mathbf{a}$ as defined in first part, for every index $n$ such that $n > N$

Combined the definition ( with annotations in brackets ) is: Sequence $\{a_n \}_{n=1}^{\infty}$ converges to $\mathbf{a}$ iff for every $\varepsilon > 0$ ( for as close as we would like to get to the number we are converging to ) there exists some $N > \mathbb{N}$ ( there is some point far down in the sequence ) such that for every $n >N$ ( such that from this point onwards ) it holds that $|a_n - \mathbb{a}| < \varepsilon$ ( is as close as we would like it to be to the number we are converging to ).

How does this all apply to your problem now? Lets first define your sequence:

$a_n = 2 - \frac{1}{n}$

We are out to prove that this sequnce converges to 2. Now how to we do it? Lets go back to definition. First part of definition "for every $\varepsilon > 0$" , and because of this we assume we have some $\varepsilon > 0$ ( it helps to fix a numeric value for epsilon in your head, but to remember that it can be absolutely any positive numeric value ). Okay now we have our $ \varepsilon $. Next we need to find for this specific $\varepsilon$ some $N \in \mathbb{N}$ such that from this point onwards we are close to 2. Here goes how we do that:

$$|a_n - 2| < \varepsilon \implies$$ $$ |2 - \frac{1}{n} - 2| < \varepsilon \implies$$ $$ |-\frac{1}{n}| < \varepsilon$$

Now we must find some some $N \in \mathbb{N}$ such that for every $n>N$ the last line of the above equation holds. Now we also know that $n>0$ since $N \in N$, thus:

$$|-\frac{1}{n}| < \varepsilon \implies$$ $$ \frac{1}{n} < \varepsilon \implies $$ $$ n > \frac{1}{\varepsilon}$$

And now we can see, that if we choose $n > \frac{1}{\varepsilon}$ then we closer to $2$ then $\varepsilon$, thus we can choose $N \geq \frac{1}{\varepsilon}$ ( any natural number larger then inverse of $\varepsilon$ ) . Lets put our proof to practical test. Say that we would like to find an index of the sequence, such that we are within $0.001$ of $2$. According to our formula for every index $ n> \frac{1}{0.001} = 1000 $ the difference between the sequence and $2$ should be less then $0.001$. Lets take $n=2000$ which is larger then $N = 1000$ and find the sequence element:

$$a_{2000} = 2 - \frac{1}{2000} = \frac{1999}{2000}$$

And truly:

$$|\frac{1999}{2000} - 2| < 0.001$$

And thus the series converges to two.

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  • $\begingroup$ very thorough and helpfull $\endgroup$
    – K. Gibson
    Commented Jul 27, 2019 at 22:12
  • $\begingroup$ I hoped it would be. As I said it took me months to understand epsilon-delta definitions, and I understand what you are going through. $\endgroup$ Commented Jul 27, 2019 at 22:17
  • $\begingroup$ Yeah i am taking real analysis in the fall and I knew this was going to be the toughest course I have ever taken in my entire life so I started studying months ahead and its a good thing I have. $\endgroup$
    – K. Gibson
    Commented Jul 27, 2019 at 22:24
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    $\begingroup$ If you really want to study analysis, go get Spivaks Calculus. This very explanation is basically rewritten out of Spivak. Once you get the hang of the limiting processes, and you can kind of imagine what is going on, the rest comes much easier. Also do not get bogged down in formalism, there is always a way to really understand it . $\endgroup$ Commented Jul 27, 2019 at 22:28
  • $\begingroup$ Okay I will look into that. We are using William R. Wade. But I usually find it helpful to get explanations from multiple sources because one might click better. $\endgroup$
    – K. Gibson
    Commented Jul 27, 2019 at 22:36
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If I may, I would like to summarize the idea:

A sequence $x_{n}$ converges to $a$ provided that given an $\epsilon > 0$, there exists a positive integer $N$ such that whenever $n \geq N$ it follows that $|x_{n} - a| < \epsilon$.

This means that if a sequence $x_{n}$ converges, then regardless of what is happening to the terms initially, eventually all of the terms must reside in an open interval centered at $a$ of radius $\epsilon$. That is what is meant by "there exists a positive integer $N$ for which whenever $n \geq N$, it follows that $|x_{n} - a| < \epsilon$."

So, for example, consider the sequence $x_{n} = \frac{\sin n}{n}$, which converges to $0$ (though not monotonically). We can show that it converges to $a = 0$ by being given an arbitrary $\epsilon > 0$, and then proceed to determine a positive integer $N$ that works:

$$ \left| \frac{\sin n}{n} - 0 \right| = \left| \frac{\sin n}{n} \right| < \frac{1}{N} < \epsilon; $$

and so, if we take $N$ to be any integer greater than $1/\epsilon$, it follows that whenever $n > N$, we have

$$\left| \frac{\sin n}{n} \right| < \frac{1}{N} < \frac{1}{\frac{1}{\epsilon}} = \epsilon. $$

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