Fürstenberg’s proof on the infinity of primes revisited

Warning to teaching staff: summing infinite sequences of positive (>0) integers is difficult, but possible. For example \sum_{n=0}^\infty n \cdot n!=0 is not a mistake and students must get full marks. Let me show you why.

A couple of months ago I was hoping to ‘cheat’ a proof of the Erdös discrepancy conjecture by using a variant of an idea of Fürstenberg’s proof on the infinitude of primes. Remember, Fürstenberg considered the integers with a new topology. Its open sets  are \{a+n b: n\in\mathbb{Z}\} for a\in\mathbb{Z} and b>0.

This topology is metrizable. There are a couple of hand-waving arguments how this metric could look like. However, as far as I am aware of, there is so far no neat description in the literature. A couple of days ago, R. Lovas and I. Mezö have published a fairly straightforward proof that d(n,m)=\|n-m\| with \|n\|:=\frac{1}{\max\left\{k\in\mathbb{N}_{>0}:1|n,2|n,\ldots,k|n\right\}} induces the above topology.

Since \|n!\|\leq\frac{1}{n} the sequence (n!)_{n\in\mathbb{N}} converges to 0 in this topology. The partial sums of \sum_{n=0}^\infty n \cdot n! satisfy \sum_{k=0}^{n-1} k \cdot k!=n! and thus \sum_{n=0}^\infty n \cdot n!=0.

R. Lovas and I. Mezö have collected more such observations in their note. What they did not mention explicitly, but what I consider interesting is that with the above metric, the integers become an ultrametric space. Without loss of generality we assume \|m\|\leq\|n\|. Then 1,2,\ldots,\frac{1}{\|n\|} are all divisors of m and n and thus they are divisors of m+n. Therefore \|m+n\|\leq \|n\|=\max\{\|m\|,\|n\|\}. The strong triangle inequality now follows d(m,n)=\|m-l+l-n\|\leq \max\{d(m,l),d(l,n)\}.


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