Proof that the sequence $s_n = \frac{1}{n}$ converges to $0$ 
Prove that the sequence $s_n = \frac{1}{n}$ converges to $0$.

I am writing this proof in order to help other people to understand better how to prove if a sequence converges and in particular why $s_n = \frac{1}{n}$ converges to $0$. 
Please, make me know if this proof is correct and how can I improve it.
 A: Let's start by the definition of convergence of a sequence.
Definition

A a sequence $x_n$ converges to some number $a$, if for any $\epsilon >0$, there exits a number $N$, such that $n > N$ implies $|x_n - a| < \epsilon$.

Now, here's my attempt to prove "easily" that the sequence $s_n = \frac{1}{n}$ really converges to $0$.
Discussion

Usually, with this kind of proofs, the best thing to do is to start from what we want to obtain, which is that for $n > N$ (for some $N$), we have $|s_n - 0| < \epsilon$, for any given $\epsilon > 0$.
Since $\epsilon$ can be anything greater than $0$ (even if usually is a small number), this suggests that $N$ must depend on $\epsilon$.
Notice that we want $$|s_n - 0| < \epsilon$$ or by replacing the formula of the sequence $s_n$ $$\left|\frac{1}{n} - 0\right| < \epsilon$$ or simply $$\left|\frac{1}{n}\right| < \epsilon$$
Since $n > 0$ (which is simply the index of the sequence that starts from $1$ usually), then the previous statement is equivalent to 
  $$\frac{1}{n} < \epsilon$$
Now, if we multiply both sides by $n$ and then by $\frac{1}{\epsilon}$, we obtain $$\frac{1}{\epsilon} < n$$
Summarizing, we started from what we wanted, and it seems that $n > \frac{1}{\epsilon}$, and this would imply what we want, that is $|s_n - 0| < \epsilon$. This make us think that $N$ could be greater or equal to $\frac{1}{\epsilon}$, since we want that from $n > N$ implies $|s_n - 0| < \epsilon$, and this is what we have just discovered.
Note that $N$ depends on $\epsilon$.

Proof

Given an $\epsilon > 0$. Let $N \geq \frac{1}{\epsilon}$. 
If $n > N \geq \frac{1}{\epsilon} $, then $\epsilon > \frac{1}{n} = \left| \frac{1}{n} \right| = \left| \frac{1}{n}  - 0\right|$, which is exactly what we wanted to show. So $s_n$ must converge to $0$.

A: Another way to handle it: substitute the sequence at hand and the expected limit in the definition.

The sequence $\dfrac1n$ converges to $0$, if for any $\epsilon >0$, there exits a number $N$, such that $n > N$ implies $\left|\dfrac1n - 0\right| < \epsilon$.

Indeed the inequality holds for all $n>\dfrac1\epsilon$ so that $N=\left\lfloor\dfrac1\epsilon\right\rfloor$ satisfies the definition.

The sequence $\dfrac1n$ converges to $0$, because for any $\epsilon >0$, $n > N=\left\lfloor\dfrac1\epsilon\right\rfloor$ implies $\left|\dfrac1n - 0\right| < \epsilon$.

