Which of these statements implies the other?

Question

$$\text{(i)} \forall \epsilon>0 \exists N \in \mathbb{N} \forall n \in \mathbb{N}: n > N \implies |x_n| < \epsilon $$

$$\text{(ii)} \exists N \in \mathbb{N} \forall \epsilon>0 \forall n \in \mathbb{N}: n > N \implies |x_n| < \epsilon $$

I have to give a "brief" reason.

I have said $(ii) \implies (i)$ Since we have found an $N$ that works for all $\epsilon>0$

I need to find an example of a sequence that holds for one but not the other. But I cant seem to figure one out.

Answer 1

Let $x_n=1/n$. Then (i) holds but (ii) does not: given any $N$, choosing $\epsilon=1/(N+2)$ violates (ii).