Pi is Irrational/Proof 2

Proof
For any integer $q$, let:


 * $\displaystyle A_n = \frac {q^n} {n!} \int_0^\pi \left[{x \left({\pi - x}\right)}\right]^n \sin x \, \mathrm d x $

Integration by Parts twice gives:

Writing:
 * $\left({\pi - 2 x}\right)^2 = \pi^2 - 4 x \left({\pi - x}\right)$

gives the result:


 * $(1): \quad A_n = \left({4 n - 2}\right) q A_{n - 1} - \left({q \pi}\right)^2 A_{n - 2}$

Suppose $\pi$ is rational.

Then $\pi = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.

We will deduce that $A_n$ is an integer for all $n$.

First confirm by direct integration that $A_0$ and $A_1$ are integers:


 * $\displaystyle A_0 = \int_0^\pi \sin x \, \mathrm d x = 2$


 * $\displaystyle A_1 = q \int_0^\pi x \left({\pi - x}\right) \sin x \, \mathrm d x = 4 q$

Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers.

Then by $(1)$ and the assumption that $q$ and $q \pi$ are integers, $A_k$ is also an integer.

So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction.

For $x \in \left[{0 \,.\,.\, \pi}\right]$, we have:
 * $0 \le \sin x \le 1$

and:
 * $0 \le x \left({\pi - x}\right) \le \pi^2 / 4$

hence:


 * $0 < A_n < \pi \dfrac {\left({q \pi^2 / 4}\right)^n} {n!}$

From Power over Factorial:


 * $\displaystyle \lim_{n \to \infty} \frac {\left({q \pi^2 / 4}\right)^n} {n!} = 0$

It follows from Squeeze Theorem that:


 * $\displaystyle \lim_{n \to \infty} A_n = 0$

Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$.

This contradicts that $A_n$ is an integer.

It follows that $\pi$ is irrational.