# Convergence of an infinite sum in the rationals

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 $\sum_{n=0}^\infty x^n$ converges in $\mathbb{Q}$ for $0 < x < 1$ and rational. One can perform the usual manipulation of partial sums (possible in $\mathbb{Q}$) to get $\sum_{n=0}^\infty x^n = \dfrac{1}{1-x} - \dfrac{1}{1-x}\lim_{n\to \infty} x^n$

(assuming the RHS exists) and hence it suffices to prove that $\lim_{n\to \infty} x^n = 0$. It is easy to prove (say with induction) that $x^{n+k} < x^n$ for all $k > 1$.

Then the limit is zero when for all $N\in \mathbb{N}$, we can find some $n\in \mathbb{N}$ such that $x^n < \dfrac{1}{N}$. This is close to being dual to the statement of the Archimedean property, which is of the form $\exists N,\ \dfrac{1}{N} < y$, for each positive rational $y$. 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 $x = \dfrac{1}{1+a/b}$ for positive integers $a, b$, since $x < 1$. One can prove (by induction) the estimate $(1+a/b)^n > na/b$ (Ross had this as a corollary of the binomial theorem, but there is a simple direct proof). Then we have $x^n =\dfrac{1}{(1+a/b)^n} < \dfrac{1}{na/b} = \dfrac{b}{a}\cdot\dfrac{1}{n}$. Given $N\in \mathbb{N}$, we can choose $n = bN$, so that $x^n < \dfrac{b}{abN} \leq \dfrac{1}{N}$, as needed.

What I like about this argument is that it uses nothing other than the ordered field axioms on $\mathbb{Q}$, 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 $x > 1$, there is an $n$ such that $x^n > 2$ (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 $n$, using the binomial theorem, namely $n = \lceil 1/(x-1)\rceil$ (i.e. round up to the ceiling).

## 7 thoughts on “Convergence of an infinite sum in the rationals”

1. Paige Bright says:

This is an amazing post– i really loved the induction proofs for rationals too even if a bit annoying after a bit. You might like how Casey Rodriguez did 18.100A on OCW (ocw.mit.edu). Just a small note for you, if you wanted to know, you missed some of the double dollar signs near the end and the backslash on \ceil

How do you put latex in your blogs :0 it looks greay

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1. David Roberts says:

Thanks! And I have fixed the latex issues.

I can’t remember how I got latex working, sorry! I just googled it when I set up years back, and forgot about it. It does look grey (maybe you meant ‘great’? :-), but that’s a problem with the theme not talking to the latex renderer properly, I think.

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1. Paige Bright says:

I did mean great lol but yeah it does look a little grey. I totally understand not remembering lol I’ll look more into it myself. Love your posts!

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2. Madeleine Birchfield says:

How much of analysis could actually be done solely on the rational numbers?

Stuff like Bishop continuous functions and uniformly differentiable functions I’d imagine should be possible to define over the rational numbers. However, other concepts like the Riemann integral and analytic functions I think require Cauchy completeness in the Archimedean field to make any sense.

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1. David Roberts says:

A very good question. Someone mentioned (I think on Twitter) that the Taylor series of rational functions should all be like this example (which is easy to see), but possibly also that this is the only class of power series that converges like this in the rationals, namely, if a power series converges on the rationals, then it is the Taylor series for a rational function. Not sure how one would show this.

But yes, some kind of analysis on the rationals would indeed be interesting, but since the integral of a rational function can be transcendental (eg $\arctan x$ being the indefinite integral of $1/(x^2+1)$. it looks like you’d be quite constrained. Getting uniformly continuous things to work over the rationals would mean having a common basis for a bit of both Archimedean and $p$-adic analysis.

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1. Madeleine Birchfield says:

I asked on MathOverflow about convergent power series on the rational numbers, and barring any errors in the proof, it seems that there do exist convergent power series on the rational numbers which are not the Taylor series for a rational function:

https://mathoverflow.net/questions/429977/in-the-rational-numbers-is-every-convergent-power-series-a-taylor-series-for-a

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2. David Roberts says:

Aha, that’s a nice argument. Thanks for asking that on MO.

I should also correct the last sentence in my last comment: of course you cannot get something to work in both the Archimedean and the $p$-adic worlds, since they have completely different metrics, and so uniform continuity in one isn’t going to hold in the other. So I should stick to my Archimedean intuition here, and leave the $p$-adic stuff either to the experts, or for later. 🙂

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