e is irrationalSuppose e to be rational. Then there is a representation e=nm with n,m∈ℕ∗. For the integer
(e−m∑i=01i!)m!=n⋅m!m−m∑i=0m!i!=n(m−1)!−m∑i=0(i+1)⋅(i+2)⋅…⋅m∈ℤ [+]
we calculate the following estimate using the limit of the geometric series:
0<(e−∑mi=01i!)m!=∑∞i=0m!i!−∑mi=0m!i!=∑∞i=m+1m!i!=∑∞i=m+11(m+1)⋅(m+2)⋅…⋅i≤∑∞i=m+11(m+1)i−m=∑∞i=01(m+1)i+1=1m+1⋅11−1m+1=1m<1
According to [+] this proves the existance of an integer within ]0,1[. Contradiction! |