I’m teaching an intro to analysis course this semester, and we are starting with the usual axiomatic treatment of numbers. I made a small emphasis on the rationals as a Archimedean field, and we can actually start with the analysis before we even get to talking about the real numbers. Moreover, since everything here is so close to the metal, we can be proving results at the level of using induction.
I wanted to use this blog post to record the proof using no more than the rationals (i.e. no embedding things into the real numbers), that the geometric series converges in for and rational. One can perform the usual manipulation of partial sums (possible in ) to get
(assuming the RHS exists) and hence it suffices to prove that . It is easy to prove (say with induction) that for all .
Then the limit is zero when for all , we can find some such that . This is close to being dual to the statement of the Archimedean property, which is of the form , for each positive rational . Initially I thought of trying to leverage the Archimedean property for the positive rational numbers, in a multiplicative sense (Archimedeanness makes sense for any ordered group), but I didn’t end up making this work (more on this below). Ultimately I found an argument in Kenneth Ross’ book Elementary analysis, which I simplified further to the following.
We write for positive integers , since . One can prove (by induction) the estimate (Ross had this as a corollary of the binomial theorem, but there is a simple direct proof). Then we have . Given , we can choose , so that , as needed.
What I like about this argument is that it uses nothing other than the ordered field axioms on , together with two very easy applications of induction. It’s a lovely proof to present to an undergrad class.
Returning my failed first idea at a proof, I had reduced the problem to that of showing that for every rational , there is an such that (challenge: can you leverage this fact to conclude the convergence as desired? It’s the base case of an induction proving the multiplicative Archimedean property). User @Rafi3AK on Twitter supplied an explicit estimate of the required , using the binomial theorem, namely (i.e. round up to the ceiling).