e is irrational

We have the series


Note that this is an alternating seriesMathworldPlanetmath and that the magnitudes of the terms decrease. Hence, for every integer n>0, we have the bound


by the Leibniz’ estimate for alternating series (http://planetmath.org/LeibnizEstimateForAlternatingSeries).  Assume that e=n/m, where m and n are integers and n>0.  Then we would have


Multiplying both sides by n!, this would imply


which is a contradictionMathworldPlanetmathPlanetmath because every term in the sum is an integer, but there are no integers between 0 and 1/(n+1).

Title e is irrational
Canonical name EIsIrrational1
Date of creation 2013-03-22 17:02:32
Last modified on 2013-03-22 17:02:32
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 8
Author rspuzio (6075)
Entry type Proof
Classification msc 11J82
Classification msc 11J72
Related topic LeibnizEstimateForAlternatingSeries