Irrationality of e

Below is a pretty standard proof of the irrationality of e, which depends on the series representation of e from evaluating the Taylor series expansion of exp about 0 at 1.

Theorem

e is irrational.

Lemma

Setting en=∑k=0n1k! we have that

e−en<1n⋅n!

for all n∈N.

Proof

The idea of this bound is to multiply through by n! to bound in terms of a geometric series.

e−en=∑k=0∞1k!−∑k=0n1k!=∑k=n+1∞1k!=1n!∑k=n+1∞n!k!=1n!(n!(n+1)!+n!(n+2)!+n!(n+3)!+…)=1n!(1n+1+1(n+1)(n+2)+1(n+1)(n+2)(n+3)+…)<1n!(1n+1+1(n+1)2+1(n+1)3+…)=1n!(1n+1(11−1n+1))=1n!(1n+1(n+1n))=1n⋅n!

We can use this bound to then construct a contradiction on e being rational.

Proof

From the previous part we know that e−en<1n⋅n! and hence for n>1 we have

0<n!(e−en)<1n<1.

Now, if we assume that e is rational, that is e=ab for integers a and b, we can deduce that

n!(e−en)=n!⋅ab−n!∑k=0n1k!=a⋅n!b−∑k=0nn!k!.

Now, because k<n in the right hand sum, we have that k!∣n! and thus this term is an integer. Similarly for n≥b, the left hand term is also guaranteed to be an integer. However this contradicts the previous constraint of n!(e−en) being strictly between two adjacent integers.