Proving that 1/2 is the least upper bound for $A = \{\frac{(-1)^n}{n}: n \in \mathbf{N} \backslash 0\}$

I'm trying to solve Problem 3(a) from https://pi.math.cornell.edu/~zbnorwood/3110-s19/3110-hw2.pdf (Spring 2019 class). We want to find the supremum of $A$ and to prove it as well.

(a) $A = \{\frac{(-1)^n}{n}: n \in \mathbf{N} \backslash 0\}$

I think $\sup A = 1/2$. To prove this, I need to prove that 1/2 is an upper bound and then I need to prove that it is the least upper bound.

For the first part, I was thinking to do it by contradiction:

Suppose 1/2 is not an upper bound then,$$1/2 < \frac{(-1)^n}{n} \\n/2 < (-1)^n.$$If we let $n = 1$, then we'll get 1/2 < -1 which is a contradiction so $1/2$ is an upper bound.

For the second part, I know that there are two strategies.

1. One is to assume that some other bound $m$ is the least upper bound and then show that $1/2 \leq m$.
2. The other strategy is use the epsilon lemma which states that $x = \sup A$ iff for all $\epsilon > 0$, there exist an element $a \in A$ such that $a > \sup A - \epsilon$. So I need to find an $n$ such that if we pick any $\epsilon > 0$, then we'll see that $$\frac{(-1)^n}{n} > 1/2 - \epsilon$$

I was thinking for strategy 2, we pick $n = 2$ and so$$\frac{(-1)^2}{2} > 1/2 - \epsilon \\ \frac{1}{2} > 1/2 - \epsilon $$This is true for any $\epsilon > 0$. So 1/2 is the least upper bound.

Is my approach correct? I do feel a little uncomfortable it with and I'm unsure why. I'm also studying this on the side so it's not a formal class where I can seek some validation.

Thank you