Sum of Geometric Progressions for any ordered Field Given an Ordered Field $\left(F,+,\cdot, 1, 0, <\right)$ it can be easily shown that 
$\sum_{i = 1}^{n} a\cdot r^{n} = a\cdot\left(\frac{r^{n + 1} - 1}{r - 1}\right)$
What I'm interested is in if $|r| < 1$ does that still mean that if $n \rightarrow \infty$ then $r^{n} \rightarrow 0$ and as such 
$\sum_{i = 1}^{\infty} a\cdot r^{n} = \frac{a}{r - 1}$
We can certainly say that $|r^{n + 1} | < |r^{n}| < 1$
But for an ordered field can we say it goes to $0$ as it does with Real Numbers?
Thanks!
 A: The better analogy would be the rational numbers, not real numbers. Given a geometric series of rational numbers, with the quotient strictly between $-1$ and $1$, the limit exists in the completion of rational numbers, i.e. in the real numbers.
You could think that given a geometric series in an ordered field, with quotient between $-1$ and $1$, the partial sums form a Cauchy sequence in the field, so the limit exists in the Dedekind completion of the field, but that is not true, unfortunately.
For example, you can order ${\mathbf R}(X)$ so that $X$ is infinitesimal. Then the series $\sum_{n} 2^{-n}$ does not converge, because its partial sums are not a Cauchy sequence, but merely a pseudo-Cauchy sequence: the differences between consecutive elements get smaller and smaller, but not arbitrarily small: namely, the difference is never infinitesimal.
The correct condition in an ordered field would be to ask that the quotient $q$ satisfies $\lim_{n\to \infty} q^n=0$. Indeed, if $X$ is infinitesimal then the series $\sum_{n} X^n$ coverges in $\mathbf R((X))$, the completion of $\mathbf R(X)$. But if you take $Y$ to be infinitesimal with respect to $X$, then $\sum_{n} X^n$ will not converge in $\mathbf R((X,Y))$.
Note that there is nothing about this conclusion that is unique to ordered fields. In any topological field, if you have $q$ such that $q^n\to 0$ and any $a$, then $\sum aq^n$ converges for more or less the same reasons.
Finally, note that in general, if a field satisfies the property that every $q\in (-1,1)$ satisfies $q^n\to 0$, then it must be Archimedean. I don't remember the reasons, but as far as I recall, all Archimedean ordered fields are subfields of the reals.
