# Infinite Number of Primes

#### Sat, Jul 22, 2017

The erroneous proof I hear most often is: Suppose \(P\) is a finite set that contains all the primes, then \(p^* = 1 + \prod_{p \in P} p\) is prime. Indeed, the flaw is that \(p^*\) is not necessarily prime but rather must be a multiple of some prime not in \(P\).

The previous proof is so often misquoted that I prefer the following. It relies on the fact that a finite product of sums of infinite geometric series converges, whereas the sum of the reciprocals of the natural numbers (except zero) diverges (hint: \(\int\)).

\[ \sum_{n = 1}^{\infty} \frac{1}{n} = \prod_{p~\text{is prime}} \sum_{j=0}^{\infty} p^{-j} = \prod_{p~\text{is prime}} \frac{p}{p-1}. \]