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How does one prove that $\zeta(3)$ is irrational ?

I would like to know how Apery did it. In particular how a recursion gives rise to irrationality !?

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You might want to review The Irrationality of $\zeta(3)$. Apery's Proof is on the Wiki for comparison to the other link. Here is another source for Apery's proof. Regards. – Amzoti Jan 18 '13 at 23:49
I don't know which recursion you mean but recursion can indeed give rise to irrationality. A classic example is $x_{n+1}=(x_n+2/x_n)/2$, which converges to $\sqrt 2$ for any $x_0>0$. – lhf Jan 19 '13 at 0:42
I meant the recursion Apery used (which is more complicated imho). – mick Jan 21 '13 at 16:27
For an overview see this answer to the question Apéry's constant (ζ(3)) value. – Américo Tavares Jan 21 '14 at 1:31
up vote 4 down vote accepted

The point is that if $\alpha\in\mathbb{Q}^+$ and $C>2$, there is at most a finite number of rational numbers $\frac{p}{q}$ such that: $$ \left|\alpha-\frac{p}{q}\right|\leq\frac{1}{q^C}. \tag{1}$$ So, if $\left\{\frac{p_n}{q_n}\right\}_{n\in\mathbb{N}}$ is a sequence of rational approximations of $\zeta(3)$ such that $(1)$ holds for every element of the sequence, $\zeta(3)\not\in\mathbb{Q}$. The problem boils down to finding "good" approximations for $\zeta(3)$. This can be done, more or less, by considering a sequence of integrals: $$ I_n = \int_{-1}^{1} P_n(x)\,f(x)\,dx = \zeta(3)-\frac{p_n}{q_n} \tag{2} $$ where $P_n$ is a Legendre polynomial. To estimate the size of the LHS is not a difficult task, but in order to estimate the size of $q_n$, too, it really useful to know that $q_1,q_2,\ldots$ satisfy some recursive formula. Given the formula, we may compute the magnitude of $q_n$ and hope that $(1)$ holds for some $C>2$ and for an infinite number of $n\in\mathbb{N}$, proving irrationality.

This is far from being an algorithmic approach: there are many well-known cases in which something goes wrong: the Euler-Mascheroni constant $\gamma$, the next interesting value of the $\zeta$ function, $\zeta(5)$, and so on.

Apery was very fond of his proof, and we may figure why: it is almost a miracle.

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In other words, if you can be approximated too well, you are irrational. – marty cohen Aug 26 '15 at 17:29
Nice. But I must say the comments were very helpful too. In fact imho part of the answer. Comments and answer + 1 😊 – mick Sep 18 '15 at 19:45

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