e is irrational

We have the series

e^{-1}=\sum_{k=0}^{\infty}{(-1)^{k}\over k!}

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

0<\left|\sum_{k=0}^{n}{(-1)^{k}\over k!}-e^{-1}\right|<{1\over(n+1)!},

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

0<\left|\sum_{k=0}^{n}{(-1)^{k}\over k!}-{m\over n}\right|<{1\over(n+1)!}.

Multiplying both sides by $n!$ , this would imply

0<\left|\sum_{k=0}^{n}{(-1)^{k}n!\over k!}-m(n-1)!\right|<{1\over n+1},

which is a contradiction 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