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I get what the epsilon-definition of convergences means but I still don't get how we find a fitting $N$. Let me explain what I mean with a little (simple) example:

Let $a_n = \frac{n}{n+1}$, determine the limit and prove it. So, the limit is easy to find: it's $1$. The definition I am using is: $$\forall \epsilon > 0: \exists N \in \mathbb{N} \text{ such that } \forall n \geq N,\ |a_n-a| < \epsilon.$$

To find $N$ I did the following: $$|a_n-a| < \epsilon \Rightarrow \left|\frac{n}{n+1}-1\right| < \epsilon \Leftrightarrow \frac{1}{n+1} < \epsilon \Leftrightarrow n > \frac{1-\epsilon}{\epsilon}$$

Does that mean I have to choose $N = \frac{1-\epsilon}{\epsilon}$? I've got a solution in which they take $N = \frac{2}{\epsilon}$ but I don't know why and..how.

Thanks in advance.

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    $\frac 2{\epsilon} > \frac {1-\epsilon}{\epsilon}$ is if it true for yours it is true for theirs. I think they just chose $2/\epsilon$ because it was easy. It doesn't matter how precise you get. You can pick an N a factor of a billion too large, it will still be true. – 2017-02-28

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Your reasoning looks good. You want to find $N$ such that the inequality $$\frac{1}{n+1}<\varepsilon$$ is satisfied for any $n \ge N$. You can achieve this by taking: $$N = \frac{1-\varepsilon}{\varepsilon}=\frac{1}{\varepsilon}-1$$ Taking any $N$ larger than this value, will (clearly) work as well. Note that $\tfrac{2}{\varepsilon} > \frac{1}{\varepsilon}-1$ so the solution's choice of $N$ works as well - not sure why they would come up with this value though.