I was reading this proof
Show that $e^{\sqrt 2}$ is irrational
I could understand until the last step.
$0<p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}<\frac{2}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}$
"But $\left(p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}\right)\in\mathbb{N}$ which is a contradiction for large n. Thus s is irrational."
I thought the contradiction would come up if you took the limit of $n \to \infty$, in the last inequality, but taken the limit doesn't help, as the left side goes to $0$, the middle goes to "$\infty$"$\cdot$$(p-q(\cosh \sqrt2))$, and the rigth side goes to $0$.
Also, I think the right member of the inequality is missing a $q$, so I rewrote as
$0<p-qs_n<\frac{2^n}{(2n)!} \frac{2q}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}$
And took the limit again, as $n$ aproaches infinity, so the left side goes to $0$, the right side goes to $0$, and the middle goes to $p-q(\cosh \sqrt2)$. I need to make sure that $p-q(\cosh \sqrt2)$ is greater than $0$ ? I don't understand the last step on the proof. If someone could explain to me I'll be thankful.
The thing at the center of the inequality $$ 0<p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}<\frac{2q}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}\tag1 $$ is a natural number (i.e., a positive integer) for every $n$. But the RHS tends to $0$ as $n\to\infty$, which means the RHS is less than $1$ for large enough $n$. The contradiction is that you've now identified a natural number between $0$ and $1$.
EDIT: To see why the central entity $$p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}$$ is an integer, note first that $(2n)!/2^n$ is an integer for each $n$ (an induction proof works). It follows that $s_n(2n)!/2^n$ is also an integer, since $$ s_n \frac{(2n)!}{2^n}=\sum_{k=0}^n \frac{(2n)!}{2^n}\frac{2^k}{(2k)!}=\sum_{k=0}^n{2n \choose 2k}\frac{(2(n-k))!}{2^{n-k}}. $$ Since $p$ and $q$ are assumed integers, the claim follows.