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!
On
(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.
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 $n > N$.