Let $S = \{\frac{(-1)^n}{n} : n \in \mathbb{N}\}$ and $f(x) = \exp(x)$. Prove that $f(S)$ is bounded from above and below and find its $\sup f(S)$ and $\inf f(S)$. Are these values equal to the maximum and minimum of $f$ on $S$?
My attempt: So $f(S)$ is the set $\{f(x) : x \in S\} = \{\exp(x) : x \in S\} = \{\exp(\frac{(-1)^n}{n}) : n \in \mathbb{N}\}$. To prove that $f(S)$ is bounded above, I need to show that there exists a $u \in \mathbb{R}$ such that $y \le u$ for all $y \in f(S)$. And to prove that $f(S)$ is bounded below, I need to show that there exists a $l \in \mathbb{R}$ such that $l \le y$ for all $y \in f(S)$. How can I find such $u$ and $l$? Furthermore, how can I then find the supremum and infimum and prove those values are indeed the supremum and infimum? Likewise for the maximum and minimum.