Help gaining a basic understanding of sequences in $\mathbb{R}$ 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.
 A: 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.
A: 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
$$\left| \frac{\sin n}{n} \right| < \frac{1}{N} < \frac{1}{\frac{1}{\epsilon}} = \epsilon. $$ 
