The sum after Leibniz's, made geometric

Turning a slow series geometric took Leibniz’s series,

$$1 - \frac13 + \frac15 - \frac17 + \cdots = \frac{\pi}{4},$$

which needs five billion terms for ten decimal places, and rewrote it with Euler’s transformation into a series whose terms halve at every step. The lesson was that a slow series and a fast one can carry the same information, and that the fast one is a rearrangement of the slow one’s own terms. It ended with the obvious next member of the family. Square the odd numbers and keep the alternating signs:

$$G = 1 - \frac19 + \frac1{25} - \frac1{49} + \cdots = 0.915965594177\ldots$$

This is Catalan’s constant, named after Eugène Catalan, who studied it in the 1860s. It converges as slowly as Leibniz’s series, it turns up in more places than its obscurity suggests, and its nature is unknown: nobody knows whether it is a fraction. This essay makes its series geometric in two different ways, uses the faster one to compute a thousand decimals, and then asks what a thousand decimals can and cannot settle about the question that matters.

Why the defining series is slow #

The terms of $1 - 1/9 + 1/25 - \cdots$ are $1/(2n+1)^2$, and an alternating series whose terms decrease to nought has an error after $n$ terms no larger than the next term. Here the next term after sixty is $1/121^2$, about $7 \times 10^{-5}$, and the actual error is about half that: four correct decimals for sixty terms of work. Each further correct decimal needs the term count multiplied by about $\sqrt{10}$, so twenty decimals would need around a billion billion terms. As a means of computing the number the series is useless.

The reason is visible in a single number: the ratio of each term to the one before. For $1/(2n+1)^2$ that ratio is $((2n-1)/(2n+1))^2$, which tends to 1. A series is measured against the geometric series, and a series whose ratio tends to 1 is as far from geometric as a convergent series can be. Every method of acceleration is, in one way or another, a method of producing a new series with the same sum and a ratio below one.

Euler’s transformation: a ratio of one half #

Euler’s transformation applies to any alternating series $a_0 - a_1 + a_2 - \cdots$ whose terms are smooth enough. It replaces the terms by their repeated differences:

$$a_0 - a_1 + a_2 - \cdots = \frac{a_0}{2} + \frac{a_0 - a_1}{4} + \frac{a_0 - 2a_1 + a_2}{8} + \cdots,$$

with the $k$-th numerator the $k$-th difference of the sequence, divided by $2^{k+1}$. For terms like $1/(2n+1)^2$ the differences shrink only slowly, so the denominators do most of the work, and the new series’ terms shrink by a factor approaching two — a geometric series with ratio one half in all but name. The figure’s middle curve settles at $0.496$ after forty terms, and the hero shows the result: after sixty terms of the transformed series the error is about $10^{-19}$, each term adding $\log_{10} 2 \approx 0.3$ of a decimal.

This is exactly what the transformation did to Leibniz’s series, and it works on Catalan’s constant for the same reason: the terms come from a function that is smooth and completely monotone, and the differences of such a function never grow. It also shows the limit of the method. A ratio of one half is the best this transformation gives for any series of this shape, so a thousand decimals would need over three thousand terms, each a difference of thousands of numbers.

Ramanujan’s identity: a ratio of one quarter #

A better series comes from a different idea. Among the formulas Srinivasa Ramanujan recorded in his notebooks is

$$G = \frac{\pi}{8} \ln\big(2 + \sqrt3\big) + \frac38 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2 \binom{2n}{n}}.$$

The sum on the right has terms with a central binomial coefficient in the denominator, and $\binom{2n}{n}$ grows like $4^n/\sqrt{\pi n}$, so the terms shrink by a factor tending to four. The third curve in the ratio figure settles at one quarter, and in the hero the error falls by $\log_{10} 4 \approx 0.6$ of a decimal per term, reaching $10^{-40}$ after sixty terms.

The identity splits $G$ into something known and something fast. The first part, $\pi/8$ times the logarithm of $2 + \sqrt3$, is about $0.517$ and can be computed to any accuracy by standard series. The second part starts at $3/8 = 0.375$ and its later terms fall away quickly — the sixth is already about $0.000012$ — so the series does almost all of its work in its first few terms and refines for ever after. Identities of this shape come from writing $G$ as an integral, $G = \int_0^1 \arctan(x)/x \, dx$, changing variables, and expanding the new integrand in a series whose coefficients are reciprocals of central binomial coefficients. The same device gives $\zeta(2) = 3\sum_{n \ge 1} 1/\big(n^2 \binom{2n}{n}\big)$, and it gave Apéry the series $\zeta(3) = \tfrac52 \sum_{n \ge 1} (-1)^{n+1}/\big(n^3 \binom{2n}{n}\big)$ that started his proof.

A thousand decimals #

With a ratio of a quarter, a thousand decimals need about $1000/0.6 \approx 1{,}700$ terms. The figures compute them in exact integer arithmetic: every quantity is a whole number standing for itself divided by $10^{1030}$, with $\pi$ from Machin’s formula, $\ln(2 + \sqrt3)$ from the series for the inverse hyperbolic tangent, and the binomial terms built by updating the previous term with one multiplication and one division. Ramanujan’s series needs 1,703 terms before its terms fall below the last digit carried, and the result agrees with the published decimals of Catalan’s constant at every place checked.

The thousand digits look like a thousand digits from a fair ten-sided die. The counts range from 78 sixes to 124 sevens, and the chi-squared statistic for the whole table is 14.6 on nine degrees of freedom, a value that a genuinely random sequence exceeds about one time in ten. That says nothing about whether $G$ is a fraction. A fraction with a denominator in the hundreds of digits has a decimal expansion that repeats only after hundreds of digits, and its first thousand would pass this test just as well. Digit frequencies are a statement about normality, which is a much stronger property than irrationality and is not known for any classical constant.

What the continued fraction can settle #

The right instrument for the fraction question is the continued fraction. Every number has one, $G = [0; 1, 10, 1, 8, 1, 88, 4, 1, 1, 7, 22, \ldots]$, and a number is a fraction exactly when its continued fraction stops. A thousand decimals do not determine the continued fraction completely, but they determine an initial part of it with certainty: the decimals pin $G$ inside an interval of width $10^{-1000}$, and every partial quotient that is the same for both ends of that interval is a partial quotient of $G$. For these decimals the two ends agree for 967 partial quotients.

The quotients follow the law that almost every number obeys. Carl Friedrich Gauss conjectured and Rodion Kuzmin proved in 1928 that for almost every real number the proportion of partial quotients equal to $k$ tends to $\log_2(1 + 1/(k(k+2)))$: about 41.5% ones, 17% twos, 9.3% threes, and a long tail. Catalan’s constant has 41.5% ones among its first 967, and its other shares are within the fluctuations a sample of that size allows. Its largest quotients so far are 1,102, 568 and 476 — the occasional very large quotient is also typical, since the expected count of quotients above $m$ falls only like $1/m$.

The continued fraction then answers the fraction question as far as it can be answered by computation. If $G$ were $p/q$, its continued fraction would stop, and $p/q$ would be its last convergent. The convergents of $G$ are known for 967 steps, and their denominators grow steadily.

The 967th denominator has 500 digits. So any fraction equal to Catalan’s constant has a denominator of at least 500 digits. The figure’s straight line is Paul Lévy’s theorem of 1936, that for almost every number the $n$-th denominator grows like $e^{n\pi^2/(12 \ln 2)}$, about $3.2758^n$; the measured rate for $G$ is 3.2890. Aleksandr Khinchin’s theorem of 1934 says that for almost every number the geometric mean of the partial quotients tends to $2.6855\ldots$; for $G$’s first 967 it is 2.7078. Catalan’s constant behaves, as far as anyone can compute, like a number picked at random.

What each digit costs #

The three series can be compared by the work a thousand decimals would take, and the comparison is more lopsided than the ratios suggest. The defining series would need around $10^{500}$ terms, a number with no physical meaning. Euler’s transformation needs about 3,300 terms, but its $k$-th term is the $k$-th difference of the original sequence, a combination of $k + 1$ numbers with binomial weights, so the whole computation is millions of multiplications of thousand-digit numbers. Ramanujan’s series needs 1,703 terms, and each term is obtained from the previous one by multiplying by $(n+1)^2$ and dividing by $(2n+1)(2n+2)$ — two operations on one large number. The constant parts cost about as much again: Machin’s formula for $\pi$ and the inverse hyperbolic tangent for the logarithm are themselves geometric series, with ratios $1/25$ and $1/3$.

So the point of a geometric rearrangement is not only the ratio but the shape of each term. A term that can be updated from its predecessor in a fixed number of operations makes the cost of $D$ digits proportional to $D$ terms times the cost of arithmetic on $D$-digit numbers. That is the scheme every record computation of a classical constant uses, organised so that the large multiplications are few. The figures do the arithmetic plainly, with no such organisation, and a thousand decimals of Catalan’s constant still take a fraction of a second.

A pattern would have been a proof #

The continued fraction of $G$ shows nothing, and it is worth being exact about what showing something would have meant. The continued fraction of $e$ is $[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots]$, with the even numbers marching through it in a fixed pattern. A pattern that can be proved to continue is a proof of irrationality, since a continued fraction that provably never stops belongs to an irrational number, and Euler proved $e$ irrational exactly that way. The square root of any non-square has a continued fraction that repeats, and the repetition is again a proof. Catalan’s constant offers neither: its first 967 quotients — 1, 10, 1, 8, 1, 88, 4, 1, 1, 7, 22 and onward — have no visible regularity, which is what a typical number’s look like, and a number whose quotients look like chance gives a proof nothing to hold on to.

The convergents still say something quantitative. Each convergent $p/q$ satisfies $|G - p/q| < 1/q^2$, and the large quotients mark the exceptionally good ones: the quotient 88 in fifth position means the convergent before it is within about $1/(88q^2)$ of $G$. How close a fraction can get to an irrational number is controlled by these quotients, and for almost every number the quotients are unbounded but rarely large. The occasional 1,102 is part of that pattern, not a sign of anything special.

Leibniz’s series and its sequel #

The two series of this essay are the first two members of a family. The Dirichlet beta function is

$$\beta(s) = 1 - \frac{1}{3^s} + \frac{1}{5^s} - \frac{1}{7^s} + \cdots,$$

so that Leibniz’s series is $\beta(1) = \pi/4$ and Catalan’s constant is $\beta(2)$. The family’s odd members are all known: $\beta(3) = \pi^3/32$, $\beta(5) = 5\pi^5/1536$, and in general $\beta(2k+1)$ is a rational multiple of $\pi^{2k+1}$. The even members — $\beta(2)$, $\beta(4)$, and on — have no such formula, and none of them is known to be irrational.

That is the exact mirror image of the zeta function. For $\zeta(s) = 1 + 1/2^s + 1/3^s + \cdots$ the even values are rational multiples of powers of $\pi$, $\zeta(2) = \pi^2/6$ and onward, and the odd values are the mysterious ones. Roger Apéry proved in 1978 that $\zeta(3)$ is irrational by finding a series for it that converges geometrically, with a ratio so small that the convergents approach faster than any fraction could; nobody has done the same for $\zeta(5)$, or for $\beta(2)$. The nearest result is Tanguy Rivoal and Wadim Zudilin’s of 2003: at least one of the seven numbers $\beta(2), \beta(4), \ldots, \beta(14)$ is irrational. Which one, the proof does not say.

The connection runs through the reason the known values are known. Both families are values of functions built from characters — $\zeta$ from the trivial one and $\beta$ from the character that is $+1$ on numbers one more than a multiple of four and $-1$ on those one less — and Euler’s method for $\zeta(2)$, extended, evaluates each function exactly at the arguments where its character’s parity matches the argument’s. The trivial character is even and gives the even zeta values; the other is odd and gives the odd beta values. Catalan’s constant sits at the argument where the parity is wrong, which is where no exact evaluation is known for either family.

Where the constant turns up #

Catalan’s constant is not a curiosity of series. It is the growth rate of a basic counting problem: the number of ways to tile a $2m \times 2m$ square with dominoes grows like $e^{4Gm^2/\pi}$, so that each square of a large board contributes $G/\pi \approx 0.2916$ to the logarithm of the count. That was found in 1961 by Pieter Kasteleyn and independently by Harold Temperley and Michael Fisher, from a determinant whose signs make it count tilings, and the same constant governs the frozen corners of large random tilings of the Aztec diamond. It also appears as the integral of $\arctan(x)/x$ from 0 to 1, as $-\int_0^{\pi/4} \ln \tan\theta \, d\theta$, and in hyperbolic geometry, where the space around the Whitehead link — two loops, one passed through the other with a twist — has volume exactly $4G$. A number that turns up in tiling, integration and geometry, and whose arithmetic nature is unknown, is exactly the kind of number for which a fast series is worth having.

What the computations cannot show #

A thousand decimals settle nothing about irrationality. They show that a fraction equal to $G$ would need a denominator of at least 500 digits, and the decimals known by record computations — billions of them — push that bound to billions of digits. No finite computation can turn such a bound into a proof that no fraction works, because there is always a larger denominator left to try.

The Gauss–Kuzmin, Khinchin and Lévy laws are about almost every number, not about this one. They hold for all real numbers outside a set of length nought, and Catalan’s constant could lie in that set. The agreement in the figures is evidence that it does not, of the same quality as the digit counts: consistent, suggestive, and proving nothing. No classical constant other than ones with explicitly known continued fractions, such as $e$, has had any of these laws proved for it.

Ramanujan’s identity is used, not proved, in the figures. The computation agrees with the published decimals of $G$ to forty places, a check the figures make, and the identity has proofs in the literature — but the agreement is not a proof, and the essay’s argument about speed would hold for any correct identity of this shape.

The rate of a quarter is not the end. Faster series for Catalan’s constant exist, with ratios of $1/16$ and smaller, and record computations use them with binary splitting, a way of organising the arithmetic so that a sum of a million terms costs little more than multiplying a few very large numbers.

Still open: whether the sequel is a fraction #

Is Catalan’s constant irrational? Nobody knows, and the reason is specific. Every proof that a classical constant is irrational has produced, by some means, a sequence of fractions $p_n/q_n$ approaching the constant faster than $1/q_n$ — fast enough that a fraction $a/b$ equal to it would force a whole number strictly between nought and one. Apéry’s series for $\zeta(3)$ did exactly that. For $G$ the series in this essay converge geometrically, but the fractions they produce have denominators that grow much faster than their accuracy improves. Ramanujan’s series is no use here at all, since its first part involves $\pi$ and a logarithm and its partial sums are not fractions. Euler’s transformed series does give fractions: its $n$-th partial sum is accurate to about $2^{-n}$, and its denominator involves $2^n$ times the square of the least common multiple of the odd numbers up to $2n + 1$, which is around $e^{4n}$. Accuracy of $2^{-n}$ against a denominator of $e^{4n}$ is a race lost by a wide margin. Whether some series for Catalan’s constant wins it, as Apéry’s did for $\zeta(3)$, is exactly the open question; finding one would very likely settle the problem, and nobody has found one.