## 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 $d(m,n)=\|m-l+l-n\|\leq \max\{d(m,l),d(l,n)\}$.