# How can this step in the proof that pi is irrational be explained?

I was going through a proof of the irrationality of $\pi$ and came across this step:

$$\frac{p^{2n+1}}{q2^{2n}n!} < 1$$ for sufficiently large n, with $p,q$ being positive integers.

This fact was given. I tried to prove it for myself, but didn't manage to get it. Could someone please give me some hints on how to obtain this result? Thanks in advance!

• Try proving $n! \geqslant \left(\frac{n}{4}\right)^n$. – Daniel Fischer Oct 4 '13 at 9:40
• Thanks for your reply! I'd guess that the best way to prove tat would be induction on n? But why does this help? Thanks again. – Joe S Oct 4 '13 at 9:45
• Not sure if induction is the best way to prove it, but certainly a good way. It helps because you have a $2^{2n}n! = 4^n n!$ in the denominator. You can then replace it by $n^n$. – Daniel Fischer Oct 4 '13 at 9:50
• That seems a bit more roundabout than just proving $n!$ dominates ordinary exponential functions. – user14972 Oct 4 '13 at 9:56
• I understand what you have said, but I'm still not sure how we can say that the expression is less than 1, that is, saying that: $$\frac{p^{2n+1}}{qn^n} < 1$$ for sufficiently large n. – Joe S Oct 4 '13 at 9:57

Let us define a sequence $\{a_{n}\}$ by $$a_{n} = \frac{p^{2n + 1}}{q(2^{2n}n!)}$$ then we can see that $$\frac{a_{n + 1}}{a_{n}} = \frac{p^{2n + 3}}{q\{2^{2n + 2}(n + 1)!\}}\frac{q(2^{2n}n!)}{p^{2n + 1}} = \frac{p^{2}}{4(n + 1)}$$ so that the ratio $a_{n + 1}/a_{n}$ tends to $0$ as $n \to \infty$. It follows by the ratio test that the series $\sum a_{n}$ is convergent and hence its $n^{\text{th}}$ term $a_{n}$ tends to zero as $n \to \infty$. It is now clear that we can find a positive integer $N$ such that $a_{n} < 1$ for all $n > N$.
(As a variation on Paramanand Singh's answer) We recognize the exponential series $$\sum_{n=0}^\infty \frac{p^{2n+1}}{q2^{2n}n!}=\frac pq\sum_{n=0}^\infty\frac{(p^2/4)^n}{n!}=\frac pq e^{p^2/4}.$$ As the exponential series converges everywhere, we conclude that the summands tend to $0$, i.e. $\frac{p^{2n+1}}{q2^{2n}n!}\to 0$ and especially $\left|\frac{p^{2n+1}}{q2^{2n}n!}\right|<1$ for $n$ sufficiently large.