Calculating $\inf \left\{\exp((-1)^nn)\right\}$ via the $\epsilon$-definition? 
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 $\epsilon > 0$. 
 A: 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 $e^{(-1)^n n} > 0$.
A: What you have is half of the definition of the supremum, so let's start there.
"Infimum" and "supremum" are terms for the greatest lower bound and least upper bound of a set.  To keep the terms straight, remember infimum/supremum $\Leftrightarrow$ inferior/superior $\Leftrightarrow$ below/above; so "infimum" has to be a lower bound and "supremum" has to be an upper bound.
So what does "$w$ is the greatest lower bound of $S$" actually mean?


*

*First, $w$ has to be a lower bound of $S$; in other words, $w\le x$ for every $x\in S$.

*Second, any greater number ($w+\varepsilon$ for any $\varepsilon>0$) is not a lower bound of $S$; for every $\varepsilon>0$, there is some $x\in S$ where $x < w+\varepsilon$.


(I've never had much luck memorizing $\varepsilon$-$\delta$ definitions directly myself; understanding the ideas and generating definitions on the fly has always been much more effective.)
On the specific set $S=\{1, e^{-1}, e^2, e^{-3}, e^4, e^{-5}, \ldots \}$, your intuition that $\inf S = 0$ and your idea of using logarithms to prove it are both correct, but you need to consider $\log\varepsilon$ for arbitrarily small (but positive!) values of $\varepsilon$.
Hope this helps!
