The race that makes ζ(3) irrational

A tail too small to be a whole number proved that $e$ is not a fraction by producing a quantity that would have to be a positive whole number if $e$ were a fraction and showing that it is less than one. An integral that cannot be a whole number did the same for $\pi$ with an integral, and ended on the number that held out longest against the method: $\zeta(3) = 1 + 1/8 + 1/27 + 1/64 + \cdots$, the sum of the reciprocal cubes. Euler had found the sum of the reciprocal squares in 1735, $\pi^2/6$, and the reciprocal cubes resisted everything for two hundred and forty years.

In June 1978 Roger Apéry announced a proof that $\zeta(3)$ is irrational, and the audience did not believe it. The proof rested on two sequences defined by a recurrence nobody had seen, on claims about them that Apéry stated without explanation, and on arithmetic that looked miraculous. Within a few months Henri Cohen, Hendrik Lenstra and Alf van der Poorten had checked every step, and van der Poorten’s account of the episode, “A proof that Euler missed”, made the argument famous.

This essay runs the argument as a computation, in exact rational arithmetic, to see what the miracle consists of. It turns out to be a race between two numbers — one growing and one shrinking — and the first figure is the race.

The shape of every such proof #

Every proof of this kind has the same skeleton. Suppose the number $x$ is a fraction $p/q$. Find whole numbers $A_n$ and $B_n$ such that $A_n x - B_n$ is never zero. If $x = p/q$, then $q(A_n x - B_n) = A_n p - B_n q$ is a whole number, and since it is not zero it is at least one in size. So $|A_n x - B_n| \ge 1/q$ for every $n$. If one can also show that $A_n x - B_n$ tends to nought, the assumption is contradicted, and $x$ is irrational.

The difficulty is entirely in finding $A_n$ and $B_n$. They must be whole numbers, and the combination $A_n x - B_n$ must be small — smaller than any fixed $1/q$ — while not being exactly zero. For $e$ the series supplies them: $A_n = n!$ and $B_n$ the partial sum times $n!$. For $\pi$ an integral against a polynomial does. For $\zeta(3)$ nothing obvious does, because its series has denominators $k^3$ that do not cancel in any helpful way, and the partial sums, multiplied out, have denominators far too large.

Apéry’s contribution was a pair of sequences in which the smallness and the wholeness are both available, though neither is obvious. The smallness costs nothing: the gap between the two sequences is tiny. The wholeness costs a multiplier, and the race is between how fast the multiplier grows and how fast the gap shrinks.

Two sequences and one recurrence #

Apéry’s sequences both satisfy the recurrence

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

Started from $a_0 = 1$, $a_1 = 5$, it produces 1, 5, 73, 1445, 33001, 819005, … — and a first surprise is that these are all whole numbers, although the recurrence divides by $n^3$ at every step. They are the sums $a_n = \sum_k \binom{n}{k}^2 \binom{n+k}{k}^2$, and the computation behind the figures checks that identity for every $n$ up to sixty. Started from $b_0 = 0$, $b_1 = 6$, the same recurrence produces fractions: 0, 6, 351/4, 62531/36, 11424695/288, …

The ratio $b_n/a_n$ approaches $\zeta(3) = 1.2020569031\ldots$ astonishingly quickly: 1.2, then 1.20205, then 1.2020569, gaining more than one and a half decimal places at every step, so that by $n = 6$ it is right to eleven places. The reason is that $a_n$ grows like $(1 + \sqrt 2)^{4n}$, about 34 times per step, and the gap $a_n \zeta(3) - b_n$ shrinks like $(\sqrt 2 - 1)^{4n}$, about 34 times per step the other way, so the ratio’s error shrinks like $(\sqrt2 - 1)^{8n}$.

The second surprise is in the denominators of $b_n$. Multiply $b_n$ by $2\,\mathrm{lcm}(1, 2, \ldots, n)^3$ — twice the cube of the least common multiple of the numbers up to $n$ — and the result is always a whole number. That is the statement the audience in 1978 found hardest to accept, and it is checked here for every $n$ to sixty. With it, $A_n = 2\,\mathrm{lcm}(1, \ldots, n)^3 a_n$ and $B_n = 2\,\mathrm{lcm}(1, \ldots, n)^3 b_n$ are whole numbers, and the skeleton applies.

The race #

Now the skeleton needs $A_n \zeta(3) - B_n$ to tend to nought, and that is a product: the multiplier $2\,\mathrm{lcm}(1, \ldots, n)^3$ times the gap $a_n \zeta(3) - b_n$. The first figure plots both and their product on a logarithmic scale. The multiplier grows: the least common multiple of the numbers up to $n$ is about $e^n$ by the prime number theorem, so its cube is about $e^{3n}$, a factor of ten every 0.79 steps on the measured stretch. The gap shrinks like $(\sqrt 2 - 1)^{4n} = e^{-3.5255n}$, a factor of ten every 0.65 steps. Since $3.5255 > 3$, the gap shrinks faster than the multiplier grows, and the product falls — by a factor of ten every three and a half steps, to below $10^{-19}$ at $n = 60$.

So for every fraction $p/q$, the product would have to stay above $1/q$, and it does not: whatever $q$ is, the product eventually falls below it. The gap is also never exactly zero: it equals a positive integral, as the account of where the sequences came from explains below. And that is the entire proof. All the difficulty lies in the two surprising facts — the whole-number $a_n$ and the bounded denominators of $b_n$ — and in the exponent $3.5255$ beating $3$.

The margin is about seventeen and a half per cent, and it is worth seeing how narrow that is. If the least common multiple grew like $e^{1.18n}$ instead of $e^n$, the multiplier would grow faster than the gap shrinks and the argument would prove nothing. The race is won, but not by much, and nobody has found a way to widen it for $\zeta(3)$ or to run an equivalent race for its neighbours.

The same route for ζ(2) #

Apéry gave the same kind of argument for $\zeta(2) = \pi^2/6$, which was already known to be irrational because $\pi$ is. The recurrence is shorter,

$$n^2 u_n = (11n^2 - 11n + 3)\,u_{n-1} + (n-1)^2 u_{n-2},$$

with whole numbers $a_n = \sum_k \binom{n}{k}^2 \binom{n+k}{k}$ and fractions $b_n$ whose denominators divide $\mathrm{lcm}(1, \ldots, n)^2$.

Here the multiplier grows like $e^{2n}$ and the gap shrinks like $\varphi^{-5n} = e^{-2.406n}$, where $\varphi$ is the golden ratio — a second, independent appearance of a famous algebraic number in what is a statement about a sum of reciprocals. The product falls, by a factor of ten every four and a half steps, and the margin is about twenty per cent. The figure confirms that both the growth and the decay match their predicted rates: the gap’s measured slope for $\zeta(3)$ is within one per cent of $4\log_{10}(1 + \sqrt 2)$, and the same check holds for $\zeta(2)$.

That $\zeta(2)$ and $\zeta(3)$ yield to the same construction, with a square and a cube in the recurrence and in the multiplier, suggested to everybody that $\zeta(5)$ would follow with fifth powers. It has not. Every attempt at a similar pair of sequences for $\zeta(5)$ produces a gap that shrinks too slowly to beat the multiplier $\mathrm{lcm}(1, \ldots, n)^5 \approx e^{5n}$, and the race is lost.

Why the even values are easy and the odd ones are not #

The reciprocal squares, fourth powers and all even powers have closed forms: $\zeta(2) = \pi^2/6$, $\zeta(4) = \pi^4/90$, and in general $\zeta(2k)$ is a rational multiple of $\pi^{2k}$. Where the coefficients come from found the first of these from the Fourier series of a simple function, and the others follow the same way. Each even value is therefore irrational, indeed transcendental, as soon as $\pi$ is, and nothing like Apéry’s race is needed for any of them.

For the odd values no closed form is known, and none is expected. The Fourier argument produces even powers because squaring a coefficient is what Parseval’s identity does, and odd powers have no such mechanism behind them. So each odd value has to be attacked on its own, and the attack has to manufacture its whole-number sequences from scratch. That $\zeta(3)$ yielded at all is the anomaly; that $\zeta(5)$ has not is what anyone would have expected before 1978.

The contrast also explains why Apéry’s $\zeta(2)$ proof was a curiosity rather than a result: it proved something already known, by a method that was new. Its value is the one shown in the second figure — the same structure, a square where the cube was, with its own algebraic growth rate — which is the evidence that the $\zeta(3)$ proof is not a one-off coincidence but an instance of something.

The skeleton, number by number #

Set beside the other irrationality proofs in this collection, the skeleton is always the same and the materials are always different. The square that cannot shrink proved $\sqrt2$ irrational by descent, where the whole numbers are the sides of ever smaller squares; which roots refuse to be fractions generalised it to every root that is not whole. For $e$ the whole numbers came from factorials, and the pattern in e’s continued fraction found the same smallness written as integrals. For $\pi$, the fraction Lambert built for the tangent used a continued fraction whose partial quotients grow.

In each case a sequence of whole-number combinations of the number is made small, and the only question is how much the wholeness costs. For square roots it costs nothing; for $e$ it costs a factorial, which the series repays at once; for $\zeta(3)$ it costs $2\,\mathrm{lcm}(1, \ldots, n)^3$, and the repayment is the exponent $3.5255$ against $3$. The proofs differ in how much margin they have, and Apéry’s has less than any of the classical ones — which is why it came last.

The least common multiple is the opponent #

The growth of $\mathrm{lcm}(1, \ldots, n)$ is the growth of Chebyshev’s function $\psi(n)$, the sum of $\log p$ over all prime powers $p^k \le n$, since the least common multiple contains each prime to the highest power not exceeding $n$. The prime number theorem says $\psi(n)/n \to 1$, and the figure shows it creeping up from below to 0.997 at $n = 1000$. Apéry’s argument for $\zeta(3)$ needs this ratio to stay below $\tfrac{4}{3}\log(1 + \sqrt2) = 1.1752$, and for $\zeta(2)$ below $\tfrac52 \log\varphi = 1.2030$.

So the proof uses the distribution of the primes, and in a weak form: an upper bound $\psi(n) \le (1 + \varepsilon) n$ for a modest $\varepsilon$ suffices, and that much was known to Chebyshev in the 1850s, long before the prime number theorem. The primes enter because the denominators of $b_n$ are built from them, and the least common multiple is the smallest number divisible by everything that can appear. A sharper analysis of exactly which primes divide the denominators of $b_n$ — some do not, to the full power — gives a slightly smaller multiplier and a slightly wider margin, which is how the best later versions of the argument improve the constants.

Bad approximations, good enough #

The proof produces fractions $B_n/A_n$ that approach $\zeta(3)$, and it is natural to expect that they must be unusually good approximations. They are not.

The best rational approximations to any irrational number are the convergents of its continued fraction, and they are correct to about twice as many digits as their denominators have. Apéry’s fractions, written over the denominators $A_n$ that make the proof work, are correct to only about 1.14 times as many digits as their denominators have — far worse than the convergents. A fraction with a 100-digit denominator from Apéry’s sequence is right to about 114 digits; a convergent with a 100-digit denominator is right to about 200.

The proof does not need good approximations, and this is the conceptual point that separates it from the approximation arguments of approached too fast to be algebraic. What it needs is a family of fractions whose denominators are known — so that their size can be bounded in advance — and whose errors fall faster than one over the denominator. The convergents are better approximations but nobody can predict their denominators, so they prove nothing; Apéry’s fractions are worse but completely explicit, and that is what makes them useful. The same quantity measures how far from rational the number is shown to be: the irrationality measure that the proof yields for $\zeta(3)$ is about 13.4, meaning that $|\zeta(3) - p/q| > q^{-13.4}$ for large $q$, a bound later improved by Georges Rhin and Carlo Viola to about 5.5.

Where the sequences came from #

Apéry did not explain how he found his recurrence, and the explanations came afterwards. Frits Beukers found in 1979 that the gap is an integral: $a_n\zeta(3) - b_n$ equals, up to a constant factor, a triple integral over the unit cube of a polynomial in three variables raised to the $n$-th power, divided by a fixed expression. The integral is visibly small, because the integrand is at most $(\sqrt2 - 1)^{4n}$ everywhere, and visibly of the form whole number times $\zeta(3)$ minus a fraction with controlled denominator, because integrating powers over the cube produces exactly such numbers. In that form the proof looks like the integral proofs for $\pi$ and for $e^r$, and the miracle becomes a choice of polynomial.

Others traced the recurrence to modular forms: the generating function of the $a_n$ is related to a modular form for a congruence subgroup, which explains why the $a_n$ satisfy congruences and why their growth rate is the algebraic number $(1 + \sqrt2)^4 = 17 + 12\sqrt2$. Neither explanation shows how to find the analogous objects for $\zeta(5)$, because neither identifies which feature of the cubes made the integral or the modular form available.

What the computation does not show #

The figures compute everything they draw exactly, up to sixty terms, and check every identity the proof uses at every one of those terms. They do not prove the identities for all $n$. That $a_n$ is always the binomial sum, that $2\,\mathrm{lcm}(1, \ldots, n)^3 b_n$ is always whole, and that the gap is never exactly zero are theorems whose proofs are algebraic — the integrality of $b_n$ in particular needs an argument about the denominators of partial sums of $1/k^3$ that the original audience reasonably wanted to see.

The rates measured in the figures are also only rates on a finite stretch. That the gap shrinks like $(\sqrt2 - 1)^{4n}$ in the limit follows from the recurrence, whose characteristic roots are $(1 \pm \sqrt 2)^4$; the computation confirms the rate to within a per cent over $n$ from 30 to 60, which is evidence that the asymptotics have set in, not a substitute for them.

One more limitation is easy to overlook. The exact arithmetic makes the figures trustworthy as computations, but the value of $\zeta(3)$ they compare against is itself computed — to four hundred digits, from a rapidly converging series of Apéry’s own — and the comparison would be meaningless if those digits were wrong. They are checked against the known decimal expansion, and four hundred digits is far more than the sixty-term gaps need, which reach about $10^{-95}$; but the chain of trust runs through that series, and a reader who wants the proof rather than the picture needs the identities, not the digits.

Still open: the odd values beyond three #

Whether $\zeta(5)$ is irrational is not known. Nor is the irrationality of any individual $\zeta(2k + 1)$ for $k \ge 2$, or of Catalan’s constant, or of $\zeta(3)/\pi^3$. The progress since Apéry has come in a different form. Tanguy Rivoal and Keith Ball proved in 2000 that infinitely many of the values $\zeta(3), \zeta(5), \zeta(7), \ldots$ are irrational, and Wadim Zudilin proved in 2001 that at least one of $\zeta(5)$, $\zeta(7)$, $\zeta(9)$ and $\zeta(11)$ is — results that run races like Apéry’s but with many constants at once, so that the gap can be made small enough by spending freedom across several numbers rather than one.

Those theorems say that irrational values exist without saying which, and the obstruction is the same race. For a single odd value beyond 3, every known construction of whole-number sequences produces gaps that shrink more slowly than the least common multiple’s power grows. Whether some better sequence exists for $\zeta(5)$, or whether the method itself has reached its limit, is not known; nor is whether $\zeta(3)$ is transcendental, which Apéry’s argument, measuring only how far $\zeta(3)$ is from fractions, cannot touch.