Infinitude of Primes Via Euler's Product Formula for Pi

L. Euler published several formulas for the number $\pi.$ One of this, namely,


$\displaystyle \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \frac{13}{12} \times \frac{17}{16} \times \frac{19}{20} \times \frac{23}{24} \times \frac{29}{28} \times \frac{31}{32} \times \cdots$

could be derived from the Leibniz series (e.g. [Courant and Robbins, 441-442]),

$\displaystyle \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} +\ldots =\sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1}.$

John Arioni suggested a way of doing that that reminds of Euler's treatment of the harmonic series:

$\displaystyle \begin{align} \frac{\pi}{4} &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} +\ldots \\ &= \bigg(1-\frac{1}{3}+\frac{1}{3^2}-\ldots\bigg)\bigg(1+\frac{1}{5}+\frac{1}{5^2}+\ldots\bigg)\bigg(1-\frac{1}{7}+\frac{1}{7^2}-\ldots\bigg)\cdot\ldots \\ &= \prod_{p\equiv 3\space mod\space4}\sum_{k=0}^{\infty}\frac{(-1)^k}{p^k}\prod_{p\equiv 1\space mod\space4}\sum_{k=0}^{\infty}\frac{1}{p^k} \\ &= \prod_{p\equiv 3\space mod\space4}\frac{1}{1+1/p}\prod_{p\equiv 1\space mod\space4}\frac{1}{1-1/p} \\ &= \prod_{p\equiv 3\space mod\space4}\frac{p}{p+1}\prod_{p\equiv 1\space mod\space4}\frac{p}{p-1}, \end{align}$

and this is exactly Euler's product (*).

John further observed that the identity implies infinitude of primes, because had the number of primes been finite, the right-hand side in (*) would have been rational, while the left-hand side $(\pi/4)$ is irrational.


  1. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1966.

|Contact| |Front page| |Contents| |Algebra| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: