Update:How is this conclusion made: "it follows from claim 2 that each $g_n$ is greater than $0$ if $n$ is large enough."
I know that $c$ is a natural number,so $c^n\gt 0$, I can say as n gets bigger $f_{k+n}(x)\rightarrow 1 $. But how about ${k(k+1)\cdots(k+n-1)}$ ? Since k is a rational number it can be a negative number too ( since in claim3 it's stated as $k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\} $)
I am having trouble while reading the proof of the 3rd claim that Miklós Laczkovich made in his proof. Can someone help me
Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\})$
Claim 1: The following recurrence relation holds: $(\forall x\in\mathbb{R}):\frac{x^2}{k(k+1)}f_{k+2}(x)=f_{k+1}(x)-f_k(x).$
Claim 2: For each $x\in \mathbb R$, $\lim_{k\to+\infty}f_k(x)=1.$
Claim 3: If $x\neq 0$ and if $x^2$ is rational, then
$(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.$
Proof: Otherwise, there would be a number $y\neq 0$ and integers $b$ such that $f_{k}(x)=ay$ and $f_{k+1}(x)=by$. In order to see why, take $y=f_{k+1}(x)$, $a=0$ and $b=1$ if $f_{k}(x)=0$; otherwise, choose integers $a$ and $b$ such that ${f_{k+1}(x)\over f_{k}(x)}={b\over a}$ and define $y= {f_{k}(x)\over a}={f_{k+1}(x)\over b}$.
In each case, $y$ cannot be $0$, because otherwise it would follow from claim 1 that each $f_{k+n}(x)(n \in \mathbb N)$ would be $0$, which would contradict claim 2. Now, take a natural number $c$ such that all three numbers ${bc\over k}$, ${ck\over x^2}$ and ${c\over x^2}$ are integers and consider the sequence
$g_n=\begin{cases}f_k(x) & \text{if }n=0\\ \dfrac{c^n}{k(k+1)\cdots(k+n-1)}f_{k+n}(x) & \text{otherwise.}\end{cases}$
Then
$g_0=f_k(x)=ay\in\mathbb{Z} y \text{ and }g_1=\frac ckf_{k+1}(x)=\frac{bc}ky\in\mathbb{Z}y.$
On the other hand, it follows from claim 1 that
$\begin{align} g_{n+2}&=\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}\cdot\frac{x^2}{(k+n)(k+n+1)}f_{k+n+2}(x)\\[5pt] & =\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}f_{k+n+1}(x)-\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}f_{k+n}(x)\\[5pt] &=\frac{c(k+n)}{x^2}g_{n+1}-\frac{c^2}{x^2}g_n\\[5pt] &=\left(\frac{ck}{x^2}+\frac c{x^2}n\right)g_{n+1}-\frac{c^2}{x^2}g_n, \end{align}$
which is a linear combination of $g_{n+1}$and $g_n$ with integer coefficients. Therefore, each $g_n$ is an integer multiple of $y$. Besides, it follows from claim 2 that each $g_n$ is greater than $0$ (and therefore that $g_n ≥ |y|$) if $n$ is large enough and that the sequence of all $g_n$'s converges to $0$. But a sequence of numbers greater than or equal to $|y|$ cannot converge to $0.$
Since $f_{1/2}(π/4) = \cos(π/2) = 0,$ it follows from claim 3 that $π^2/16$ is irrational and therefore that $π$ is irrational.
I think this might clarify.
Claim 2 asserts that $\lim_{m\to\infty}f_m(x)=1$. Using the definition of limit with $\epsilon=1$, there is some integer $M$ for which $m>M$ implies $f_m(x)>1-1=0$. Thus for any fixed $k$, $f_{k+n}(x)>0$ if $n$ is large enough to ensure that $k+n>M$.
Suppose $n$ is “large enough.” The number $g_n$ is a positive multiple of $f_{k+n}(x)$, so $g_n$ is also greater than zero for such large enough $n$.
The number $g_n$ was shown to be an integer multiple of $y$, and the smallest positive integer multiple of $y$ is $|y|$, so if $g_n>0$, it must be at least $|y|$. (Any smaller positive value could not be an integer multiple of $y$.)
Now we have that $g_n>|y|>0$ for all “large enough” $n$, and so the sequence $g_n=f_{k+n}\cdot \dfrac{c^n}{k(k+1)\cdots(k+n-1)}$ cannot approach zero in the limit as $n\to\infty$. On the other hand, $\lim_{n\to\infty} f_{k+n} = 1$, and $\lim_{n\to\infty} \dfrac{c^n}{k(k+1)\cdots(k+n-1)} = 0$ (the new terms in the denominator eventually become and stay larger than $c$). Therefore $\lim_{n\to\infty} g_{n} = 0$, since $g_n$ is the termwise product of two sequences, one of which approaches $1$ and the other of which approaches zero.
It is here that a contradiction arises, proving Claim 3. The $g_n$ cannot approach zero, yet they do, so the presumption that Claim 3 was false is mistaken.