Show that the sequence converges to 0 Given a sequence $\{\frac{(-1)^n}{n}\}$ show directly from the definition that it converges to $0$. 
Definition of convergence of a sequence is:

A sequence $\{p_n\}$ converges if for every $\epsilon>0$ there exists an $N\in\mathbb N$  so that  $n\geq N \implies  d(p_n,p)<\epsilon$

My approach 
What we want to show is:

For every $\epsilon>0$ there exists an $N\in\mathbb N$ so that 
          $$n\geq N \implies  d(\frac{(-1)^n}{n},0)<\epsilon$$

After this step the solution I have stops making sense

Take $N$ so large that $N>\frac2\epsilon$. (No idea why $2$ is chosen?
  And why this works) Then $n\geq N \implies | \frac{(-1)^n}n | < \frac2n< \frac2N < \epsilon $ (No idea why this is true or logically progresses)

Edit: 
It could be 2/n because  $| \frac{(-1)^n}n | =| \frac{1}n |$ we need at least 2 so that $| \frac{1}n |< 2/n$
 A: We want to show that for any $\epsilon > 0$, there exists a $N \in \mathbb{N}$ such that for all $n \geq N$, $|\frac{(-1)^n}{n} - 0 | < \epsilon$. Noting that $|\frac{(-1)^n}{n} - 0 |  = |\frac{(-1)^n}{n} | = \frac{|(-1)^n|}{|n|} = \frac{1}{n}$, we need to choose $N$ such that $\frac{1}{n} < \epsilon$  or equivalently, $n > \frac{1}{\epsilon}$ for all $n \geq N$. One choice is $N=ceil(\frac{1}{\epsilon})+1$. Another is $n \geq ceil(\frac{2}{\epsilon})$ - the posiblities are endless.
A: The inequalities in the last step are simple to explain. First of all, you have the inequality $$\left|\frac{(-1)^n}{n}\right|<\frac2n.$$
This is quite a simple inequality, since you should know that $$\left|\frac{(-1)^n}{n}\right|=\frac1n$$ and since $\frac1n > 0$, it is clear that $\frac 2n > \frac 1n$.
The second inequality is also simple, that is, $$\frac2n \leq \frac 2N.$$ (yes, there should be a $\leq$ sign. This follows from the fact that $n\geq N$, meaning that $\frac 1n \leq \frac 1N$ (since in $\frac 1n$, you are dividing $1$ by a larger number).
The third inequality, $$\frac 2N<\epsilon,$$ comes directly from $$N>\frac2\epsilon$$ (just multiply the inequality by $\epsilon$ and divide by $N$.
