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e  is irrational


Suppose e to be rational. Then there is a representation e=nm with n,m. For the integer

(emi=01i!)m!=nm!mmi=0m!i!=n(m1)!mi=0(i+1)(i+2)m       [+]

we calculate the following estimate using the limit of the geometric series:

0<(emi=01i!)m!=i=0m!i!mi=0m!i!=i=m+1m!i!=i=m+11(m+1)(m+2)ii=m+11(m+1)im=i=01(m+1)i+1=1m+1111m+1=1m<1

According to [+] this proves the existance of an integer within ]0,1[.   Contradiction!