A hyperreal $\epsilon$ is infinitesimal if for every standard natural number $n$, either $-1/n < \epsilon$, or $\epsilon < 1/n$. (Here, our relationships are in the context of some ultrafilter on the natural numbers.)
A hyperreal $E$ is infinite if for every standard natural number $n$, either $E < -n$ or $n < E$.
A hyperreal $\rho$ is finite if there exists a standard natural number $n$ such that $-n < \rho < n$. Clearly, every infinitesimal $\epsilon$ is a finite number, since $-1 < \epsilon < 1$.
The multiplicative inverse of a finite number $\rho \geq 1$ is finite (let $n$ be the standard natural number which bounds $\rho$): \begin{align*} &\quad\; -n < \rho < n \\ &\Leftrightarrow -n < \rho \;\wedge\; \rho < n \\ &\Leftrightarrow -\frac{1}{\rho} < \frac{1}{n} \;\wedge\; \frac{1}{n} < \frac{1}{\rho} \\ &\Leftrightarrow \frac{1}{\rho} > -\frac{1}{n} \;\wedge\; \frac{1}{n} < \frac{1}{\rho} \\ &\Leftrightarrow -\frac{1}{n} < \rho < \frac{1}{n} \\ &\Leftrightarrow -1 < \rho < 1 \end{align*}
Let $-1 < \rho < 1$ be a finite number. \begin{align*} &\quad\; -1 < \rho < 1 \\ &\Leftrightarrow -1 < \rho \;\wedge\; \rho < 1 \\ &\Leftrightarrow -\frac{1}{\rho} < 1 \;\wedge\; 1 < \frac{1}{\rho} \\ &\Leftrightarrow \frac{1}{\rho} > -1 \;\wedge\; 1 < \frac{1}{\rho} \\ &\Leftrightarrow -1 < \frac{1}{\rho} < 1 \end{align*}
This does not make sense. I would have expected: $$-1 > \frac{1}{\rho} \;\vee\; 1 < \frac{1}{\rho} $$
For example, if $\epsilon$ is an infinitesimal, then for any standard natural $n$: \begin{align*} &\quad\; -\frac{1}{n} < \epsilon \;\vee\; \epsilon < \frac{1}{n} \\ &\Leftrightarrow -\frac{1}{\epsilon} < n \;\vee\; n < \frac{1}{\epsilon} \\ &\Leftrightarrow \frac{1}{\epsilon} < -n \;\vee\; n < \frac{1}{\epsilon} \end{align*} In other words, $1/\epsilon$ is an infinite hyperreal.
I am doing something incredibly silly with the case where $-1 < \rho < 1$, but I can't see it :( Help?
You made a mistake when you wrote $-1<\rho\iff-\dfrac1\rho<1$ .
That works if $0<\rho$, but if $\rho<0$ then $-1<\rho\iff -\dfrac1\rho\color{red}>1$.
(Don't forget to reverse the inequality sign when multiplying or dividing both sides by a negative number.)