Apéry’s theorem

In 1978, at a conference talk met with open disbelief, Roger Apéry announced that

$$\zeta(3) \;=\; \sum_{n\ge1}\frac{1}{n^3} \;=\; 1.2020569\ldots$$

is irrational. The proof used nothing later than the eighteenth century, and it took the audience weeks to accept that it worked.1

1

A. van der Poorten, A proof that Euler missed…, Math. Intelligencer 1 (1979) — still the best account of the episode.

The engine is one recurrence,

$$n^3 u_n \;=\; (34n^3-51n^2+27n-5)\,u_{n-1}\;-\;(n-1)^3 u_{n-2},$$

run from two seeds. With $b_0=1, b_1=5$ it generates the Apéry numbers $1, 5, 73, 1445, 33001,\ldots$ — miraculously integers, with the closed form

$$b_n \;=\; \sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}.$$

With $a_0=0, a_1=6$ it gives rationals whose denominators divide $2\,d_n^3$, where $d_n = \mathrm{lcm}(1,\ldots,n)$. Both grow like $\alpha^n$ with $\alpha = (1+\sqrt2)^4 = 33.97\ldots$, and the ratios $a_n/b_n$ converge to $\zeta(3)$ absurdly fast:

$$\bigl|\,b_n\,\zeta(3)-a_n\,\bigr| \;\sim\; C\,\alpha^{-n}.$$

The integer race

Suppose $\zeta(3) = p/q$. Then $N_n = 2\,q\,d_n^3\,\bigl(b_n\zeta(3)-a_n\bigr)$ is a nonzero integer. By the prime number theorem $d_n = e^{n(1+o(1))}$, so

$$|N_n| \;\le\; C’q\,\Bigl(\frac{e^{3}}{\alpha}\Bigr)^{\!n} \;\longrightarrow\; 0,$$

because $e^3 = 20.08\ldots < 33.97\ldots = \alpha$. A nonzero integer below 1 is the contradiction. The whole proof is that inequality: the recurrence converges just faster than the least common multiples grow, with nothing to spare.

Beukers (1979) repackaged nonvanishing and the size bounds as positivity of explicit integrals; the base case is already handsome:

$$\int_0^1\!\!\int_0^1 \frac{-\ln(xy)}{1-xy}\,dx\,dy \;=\; 2\,\zeta(3),$$

and inserting shifted Legendre polynomials shrinks it at the required $\alpha^{-n}$.2

2

The same skeleton with one fewer integral proves $\zeta(2) = \pi^2/6$ irrational — the method, not the miracle, is general.

What it did not open: no other odd zeta value has followed in forty-eight years. Rivoal (2000) showed infinitely many of $\zeta(5), \zeta(7), \ldots$ are irrational; Zudilin cut one case to the bone — at least one of $\zeta(5), \zeta(7), \zeta(9), \zeta(11)$ is irrational. Which one remains exactly as open as everything else in this corner. $\zeta(3)$ alone carries a certificate, and it is Apéry’s.