Define the infimum of a set. For $S = \left\{\exp((-1)^nn): n \in \mathbb{N}\right\}$, find $\inf(S)$ and prove your claim.
Definition: $w = \inf(S)$ iff for every $\epsilon > 0$ there's $x_{\epsilon} \in S$ such that $x_{\epsilon} < w+\epsilon$.
Claim: Since $e^{y} > 0$ for all $y \in \mathbb{R}$, $S$ bounded below by $w=0.$ We claim $0 = \inf(S)$.
I'm stuck on the proof because for constructing such $x_{\epsilon}$ I run into the problem of having $\log(-\epsilon)$ when working backwards from $e^{-2k-1}> -\epsilon$, and which doesn't make sense in $\mathbb{R}$ because $\epsilon > 0$.
Let $\epsilon > 0$, we need to show that there exists some $n$ so that: $$e^{(-1)^nn} - 0 <\epsilon$$ Clearly if $n$ is odd then, $$e^{(-1)^n n} = e^{-n} <\epsilon$$
So picking any odd $n>\log(\tfrac1\epsilon)$ will work. It follows that $0$ is a limit point of $\left\{e^{(-1)^n n} : n=0,1,\dotsc\right\}$, but to see that it is the infimum simply note that $e^{(-1)^n n} > 0$.