The irrationality of e

This is the post excerpt.

The irrationality of e can be proved using the infinite series expansion.

e=\sum_{k=0}^{\infty} \frac{1}{k!}

Assume that e is rational.i.e e=\frac{p}{q} where p \in \mathbb{Z} and q \in \mathbb{Z}-{0}.

Since e is positive and 2.5<e<3(I’m not going to prove this.It’s quite simple.\frac{5}{2}=1+1+\frac{1}{2},use this as a lower bound for e and proceed to the upper bound)

This ensures that q \neq 2 ie q \geq 3

e=\frac{p}{q}=\sum_{k=0}^{\infty} \frac{1}{k!} =\sum_{k=0}^{q} \frac{1}{k!}+\frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots

Multiplying both sides by q!;

p(q-1)!=q!(\sum_{k=0}^{q} \frac{1}{k!})+(\frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\frac{1}{(q+1)(q+2)(q+3)}+\cdots)

p(q-1)! and the sum q!(\sum_{k=0}^{q} \frac{1}{k!}) are integers.So the infinite sum A=\frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\frac{1}{(q+1)(q+2)(q+3)}+\cdots must also be an integer.

Now,we proceed to finding an upper bound or A,closing in on our contradiction.

A<\sum_{k=1}^{\infty} \frac{1}{(q+1)^k}

The infinite sum which provides the upper bound can be maximized by minimizing the value of q.Taking the minimum value of q which is 3.

A<(\sum_{k=1}^{\infty} \frac{1}{(4)^k}=\frac{1}{3}

Since A is clearly non-negative,this is a contradiction as A must be an integer.

e isn’t rational.

 

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