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?

  • 5
    $\begingroup$ $10^n/(10^n-1)$ is always in lowest terms, but what is its limit? $\endgroup$
    – MBW
    Jul 17, 2015 at 20:45
  • 4
    $\begingroup$ Irrationality of $\pi$ is somewhat tricky. I definitely recommend reading Wikipedia on the subject if you're curious! $\endgroup$
    – MBW
    Jul 17, 2015 at 20:53
  • 6
    $\begingroup$ Proving $\pi$ irrational is not something that a person would expect a person to be able to do by just puttering around like this. It's hard. $\endgroup$ Jul 17, 2015 at 20:53
  • 4
    $\begingroup$ The series for $e$ can be used to prove irrationality of $e$ in a simple way. I do not know of an analogous argument for $\pi$. $\endgroup$ Jul 17, 2015 at 21:16
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    $\begingroup$ Since we are used to having both $\pi$ and $\sqrt{2}$ in math textbooks right from class 7-8 (students of age 12-13 years) it appears that these numbers are on the same footing. No!!!! $\pi$ sits on a very very high level compared to $\sqrt{2}$. There is an easy way to define and compute $\sqrt{2}$ but no equally simpler way to define/compute $\pi$. To prove the irrationality of $\pi$ we need to use its specific properties which are bound to somewhat complicated compared to the properties of $\sqrt{2}$. $\endgroup$
    – Paramanand Singh
    Aug 23, 2015 at 5:50

1 Answer 1


See this article: Proof that $\pi$ is irrational.

$\pi$ was not proved to be irrational until the 18th century. It's not easy or it would have been done earlier. Cartwright's and Niven's proofs (are they really the same proof?) were published in the 20th century and can be followed by someone familiar only with freshman calculus, but to come up with the arguments in the first place was not easy. (Today we know of Cartwright's proof only because it appeared in an appendix of one edition of a book by Harold Jeffreys on scientific inference.)


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