real analysis - 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 $| \frac{1}n |< 2/n$

Answer

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 $N$.