real analysis - 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!

Answer

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 $n > N$.