I was trying to prove that $\pi$ is irrational, just to see if I could do it. So far, I've tried to do this by using the fact that the sum
$$S=\sum\limits_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ and arguing that $S$ is irrational instead, and thus implying that $\pi$ is also irrational.
To do this, I thought I could use partial sums $S_n$ of $S$:
$$S_n=\sum\limits_{k=1}^n \frac{1}{k^2}=\frac{A_n}{B_n}$$ where the fraction $A_n/B_n$ is written in lowest terms. We can note that the sequence $\lbrace S_n \rbrace$ is a strictly increasing sequence, with $\pi^2/6$ as its lowest upper bound, so I thought that maybe the sequences $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ both have no upper bound, and the sequences would tend to infinity, and if I were able to prove this, I thought I could use it to argue that $S$ is an irrational number, using the fact that every rational number can be written as the quotient between two finite integers.
We note that $A_n\geq B_n$ for all natural $n$, as $S_n\geq 1$, so all we would really need to prove in that case is that $\lbrace B_n \rbrace \rightarrow \infty$(if this assertion is even correct).
At first I thought that both $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ would be increasing sequences, but after checking with Maple I noticed that they weren't, sadly enough($S_9=\frac{9778141}{6350400}$ and $S_{10}=\frac{1968329}{1270080}$). However, they do indeed seem to get very large very quickly, so I'm thinking that my hypothesis about $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ is correct.
But I have trouble proving my hypothesis, and I'm kind of stuck, no knowing what to do. Is there a way to prove that $\lbrace B_n \rbrace \rightarrow \infty$? And of course, is this approach to prove that $\pi$ is irrational logically sound, or is it fundamentally flawed in some important aspect? In the latter case, what idea should I try next?
