Elementary proof that $\pi$ is irrational

Question

I'm trying to understand the first proof in this page. So we have

$$S=\frac{\pi }{4}=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}=S_{n}+R_{n}$$

where $S_{n}=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{2k-1}$ and $R_{n}=\sum_{k=n+1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$. Since the series is alternating we determine

$$0<\mid R_{n}\mid<\frac{1}{2n+1}$$.

Now we assume $S$ is rational. $S_{n}$ is clearly rational so $R_{n}=S-S_{n}$ must be a rational number. So far so good. But i didn't understand the next step. Can you explain it to me?

Answer 1

The referenced article is wrong, so wrong, my god it's wrong.

It says "For if the contrary is true, i.e. if S is rational, then since Sn is a rational fraction, the first of equations (2) says that a rational number S equals a rational number Sn plus, in view of equation (3), a non-vanishing fraction Rn, which is impossible."

This statement is absurd. Here is an obvious counterexample: $\frac12 = \frac13+\frac16$.

To explicate: The rational number $\frac12$ equals a rational number $\frac13$ plus a non-vanishing fraction $\frac16$.

Poor guy.

Answer 2

A proof that needs nothing else than elementary integral calculus was given by Niven here.

https://projecteuclid.org/download/pdf_1/euclid.bams/1183510788

Unlike the proof you post, you'll find this proof is correct. You can find a complete exposition of this proof in Ross's $\textit{Elementary Analysis}$ text.