What you want to prove is that
For all $\epsilon >0$, there exists a natural number $N$ such that, whenever $n\geq N$, we have $$\left|\frac{(-1)^n}{n}\right|<\epsilon$$
But $\left|\dfrac{(-1)^n}{n}\right|=\dfrac 1 n$.
In fact, what you want to prove is then that
For all $\epsilon >0$, there exists a natural number $N$ such that, whenever $n\geq N$, we have $$\dfrac 1 n <\epsilon$$
This is in fact a consequence of the unboundedness of the natural numbers. Particularily
For all $\epsilon >0$, there exists a natural number $N$ such that, $$\dfrac 1 N <\epsilon$$
is what we call the Archimedean property of the real numbers. Basically, if there was some $\epsilon$ such that for no natural number $n$
$$\dfrac 1 n <\epsilon$$
would hold, this would mean
$$\dfrac 1 n \geq \epsilon$$
for all natural numbers, so
$$\dfrac 1 \epsilon \geq n$$
which would mean $\dfrac 1 \epsilon$ is an upper bound for the natural numbers, which is impossible.
As Ross is suggesting, since $$n\frac 1 n>\frac 1{n+1}$$ so if there is one $N$ that works, all other naturals greater than it will also work, so you can put all of the above into a proof.