A series that waits on π

No integrand sits on the border found that there is no dividing line between convergent and divergent series: below every divergent one sits a slower-diverging one, above every convergent one a slower-converging one. It ended on a series whose side of that non-existent line nobody knows:

$$\sum_{n=1}^{\infty} \frac{1}{n^3 \sin^2 n}.$$

It is called the Flint Hills series, after a remark in a 2002 book by Clifford Pickover, and it looks harmless. The factor $1/n^3$ makes the terms small, and $\sin^2 n$ is at most one, so the terms are at least $1/n^3$ and usually close to it. The trouble is that $\sin^2 n$ is also sometimes very small, whenever the whole number $n$ is close to a multiple of $\pi$, and then the term is large. How often that happens, and how close it gets, is a question about $\pi$ rather than about series.

This essay computes the series as far as a computer comfortably can, finds exactly where its size comes from, and shows why no amount of computing will decide whether it converges.

Ten million terms #

Summed directly, the series behaves oddly. The first term is 1.41, since $\sin 1 = 0.841$. The third is 1.86, because $\sin 3 = 0.141$ is small: three is close to $\pi$. The twenty-second is 1.20, because 22 is close to $7\pi = 21.99$. And the three hundred and fifty-fifth is 24.60, because 355 is extraordinarily close to $113\pi = 354.99997$, so that $\sin 355 \approx -0.00003$ and its square is about $9 \times 10^{-10}$.

After that, nothing visible happens. The sum at ten million terms is 30.3145, and the ordinary terms, of size about $1/n^3$, contribute almost nothing beyond the first few hundred. A reader shown only this staircase would say the series obviously converges, to about 30.3, and would be making exactly the mistake the series exists to illustrate: the sum’s growth is concentrated in rare spikes, and whether there are enough large spikes further out is invisible from any finite stretch.

How the question was made precise #

The series became a standard example after Pickover posed it in 2002, and for several years its connection to $\pi$ was a heuristic: the large terms were obviously at the good fractions, so convergence obviously depended on how good those fractions could get. Max Alekseyev made the connection a theorem in 2011. His argument has two directions. If the measure is below $5/2$, the spikes at every good fraction are bounded by a shrinking power of the numerator, and the ordinary terms are controlled because $|\sin n|$ can be small only when $n$ is near a convergent’s numerator or a small multiple of one; adding the two bounds shows the sum is finite. Conversely, if the series converges, its terms go to nought, so $p^{2\mu - 5}$ cannot stay large along the convergents, and that forces the measure to be at most $5/2$.

So the series is not merely related to the irrationality measure; it is a restatement of one bound on it, with a boundary case at exactly $5/2$ where the argument does not decide. The same kind of equivalence holds for the whole family of series with other powers, each tied to its own threshold. The Flint Hills series is the member of the family whose threshold, $5/2$, sits closest to the believed value, 2, while still being far below anything proved.

Why the usual tests say nothing #

Every test for convergence taught first compares a series with a known one. Here the natural comparison is with $\sum 1/n^3$, which converges, as an endless region with a finite area showed for any power above one. But the comparison runs the wrong way: $\sin^2 n \le 1$ makes each term at least $1/n^3$, which proves nothing about the sum being finite. To bound the terms above one needs a lower bound on $|\sin n|$, and there is no constant lower bound: the values $n$ taken modulo $2\pi$ come arbitrarily close to $0$ and to $\pi$, because $\pi$ is irrational, so $\sin n$ comes arbitrarily close to nought.

The ratio test and the root test fail too, for a related reason: they look at how one term compares with the next, and the terms here jump by factors of millions at unpredictable places. Nothing that inspects the terms locally can see whether the rare large ones are rare enough. The question has to be translated into how close $n$ can come to a multiple of $\pi$, which is not a question about series at all.

How often sin n is small #

It helps to ask how the series would behave if the values $n \bmod 2\pi$ were scattered at random, which is the right model for a typical number in place of $\pi$. The values of $n$ modulo $2\pi$ do spread evenly round the circle — they are equidistributed, like the fractional parts in almost every orbit is fair — so the chance that $|\sin n| < \varepsilon$ is about $2\varepsilon/\pi$ for small $\varepsilon$.

Under that model the smallest value of $|\sin k|$ among the first $n$ values of $k$ is typically about $1/n$, and a value as small as $1/n^{2}$ turns up among them only with probability about $1/n$; in each later stretch from $n$ to $2n$ the chance of such a coincidence shrinks in step. They keep happening, more and more rarely, and the $n^3$ in the denominator easily outweighs them. The model predicts convergence with room to spare, and the spikes at $\pi$’s convergents behave exactly as the model predicts. What the model cannot supply is a proof that $\pi$ is typical in this sense, and the series is a direct test of that.

Where the spikes come from #

On logarithmic axes the terms form a band hugging the line $1/n^3$ from above, with spikes. Every spike is at a whole number $p$ for which some fraction $p/q$ is an unusually good approximation to $\pi$. Then $p - q\pi$ is small, and since $\sin p = \sin(p - q\pi)$ up to sign, $\sin p$ is about as small as the error:

$$\frac{1}{p^3 \sin^2 p} \approx \frac{1}{p^3 (p - q\pi)^2}.$$

The best approximations to $\pi$ by fractions are its convergents, the fractions produced by its continued fraction $[3; 7, 15, 1, 292, 1, 1, 1, 2, \ldots]$, and the fractions that beat every smaller one showed that they are exactly the fractions closer to the number than any with a smaller denominator. They are 3, 22/7, 333/106, 355/113, 103,993/33,102, 104,348/33,215, and on. Every spike in the figure is at one of their numerators.

The table makes the mechanism exact. The error of 355/113 is $3.0 \times 10^{-5}$, unusually small because the next partial quotient in $\pi$’s continued fraction is 292: a large partial quotient means the convergent before it is an exceptionally good approximation. That is the source of the jump. The later convergents have errors that shrink roughly like $1/p$, the ordinary rate for a number whose continued fraction has modest terms, and their contributions fall — $2.4 \times 10^{-6}$ at 103,993, $3.2 \times 10^{-7}$ at 833,719.

The exponent that decides #

The last column of the table is the quantity that decides the series. Write the error of a convergent as $|p - q\pi| = p^{1-\mu}$, defining an exponent $\mu$ for each fraction. Then the term at $n = p$ is about

$$\frac{1}{p^3 \cdot p^{2 - 2\mu}} = p^{2\mu - 5}.$$

If every good fraction has $\mu$ comfortably below $5/2$, the spikes shrink like a negative power of $p$, and since the convergents’ numerators grow at least exponentially, the spikes add to a finite amount; the ordinary terms, of size about $1/n^3$, add to a finite amount too. If infinitely many fractions have $\mu$ above $5/2$, the spikes grow without bound and the series diverges.

The largest $\mu$ that infinitely many fractions achieve is the irrationality measure of $\pi$: the supremum of the exponents for which $|\pi - p/q| < 1/q^{\mu}$ has infinitely many solutions. For a rational number it is 1; for every algebraic irrational it is exactly 2, by Klaus Roth’s theorem of 1955; and for almost every real number it is 2 as well. Approached too fast to be algebraic built numbers with measure infinity, which is how Liouville constructed the first transcendental numbers. Max Alekseyev showed in 2011 that the Flint Hills series converges if $\pi$’s measure is below $5/2$ and only if it is at most $5/2$.

On the evidence of every convergent computed, $\pi$’s measure is 2 — its continued fraction looks like a typical number’s, with occasional large terms like 292 of the frequency typical numbers have. If that is right, the series converges. But the measure is a statement about infinitely many fractions, and no finite computation of convergents bounds it.

Three numbers, three behaviours #

The dependence on the number, rather than on the series, can be made visible by replacing $\pi$. The small values of $\sin n$ are the small values of $\|n/\pi\|$, the distance from $n/\pi$ to the nearest whole number, so the Flint Hills series is essentially the series $\sum 1/(n^3 \|n\theta\|^2)$ at $\theta = 1/\pi$. Put other numbers in its place.

For the golden ratio, which fractions approximate worse than they approximate any other number, every $\|n\theta\|$ is at least a constant over $n$, the spikes are at most of size $1/n$, they sit at the Fibonacci numbers, which grow exponentially, and the sum settles at 13.78. For $1/\pi$ the sum settles, as far as anyone can compute, at 294.29. For a number constructed so that each partial quotient is the square of the denominator before it, each good fraction has exponent about 4, its spike is about $p^3$, and the sum races past $10^{10}$ within 300,000 terms and diverges.

The three sums are the same series with different numbers inside, and the whole difference is how well each number can be approximated by fractions. It is the same arithmetic that decided, in how smooth the disguise is, whether the change of coordinates for a circle map was smooth: there the divisors $e^{2\pi i k\rho} - 1$ were small when $k\rho$ was close to a whole number, and here the divisors $\sin n$ are small when $n$ is close to a multiple of $\pi$. Small divisors are the same phenomenon wherever they appear.

What is proved about π #

What has been proved is much weaker than what is believed. Kurt Mahler showed in 1953 that $\pi$’s measure is finite, at most 42, which already made $\pi$ provably not a Liouville number. The bound has been lowered by a series of increasingly sophisticated constructions of rational approximations to integrals involving $\pi$ — the same kind of integral that cannot be a whole number was used to prove $\pi$ irrational — to 7.10 by Doron Zeilberger and Wadim Zudilin in 2020.

The same bound does decide other series. With $n^u$ in place of $n^3$, the spikes are of size $p^{2\mu - 2 - u}$ and the threshold moves to $\mu < 1 + u/2$. Since $\pi$’s measure is proved to be below 7.1, the series with $u$ above 12.2 provably converge: $\sum 1/(n^{13} \sin^2 n)$ is finite, and so is every series with a higher power. The proof gives no help below that, and the gap between thirteen and three is the gap between what is proved about $\pi$ and what is believed.

Between 7.10 and 2.5 there is nothing. The series converges if the true measure is below 2.5, and the proofs cannot yet rule out that it is 7. Settling the Flint Hills series would require, at the least, bringing the bound on $\pi$’s measure down by a factor of nearly three, and nobody knows how to do that.

Why computing more terms cannot help #

The partial sums through ten million are 30.3145, and computations carried much further move only the later digits. None of that bears on convergence. The terms that could make the series diverge are spikes at the numerators of future convergents with unusually large partial quotients after them, and whether $\pi$’s continued fraction has infinitely many partial quotients large enough — growing faster than the square root of the denominators — is exactly the question of its irrationality measure.

The continued fraction of $\pi$ has been computed to billions of terms, and its partial quotients follow the statistics a random number’s would — the Gauss–Kuzmin distribution, with a term of size about $N$ appearing roughly once in every $N$ terms — which is why the measure is believed to be 2. But a statistic of billions of terms is not a statement about all of them, and a single partial quotient of astronomical size somewhere beyond reach could create a spike larger than everything before it. A cube root that looks like chance met the same wall for $\sqrt[3]{2}$: a continued fraction that behaves like a typical number’s for as far as it has been computed, and no proof that it keeps doing so.

What the figures cannot show #

The partial sums use ordinary double-precision arithmetic, which computes $\sin n$ to about sixteen significant digits for $n$ up to ten million. The smallest values of $\sin n$ in that range are about $4 \times 10^{-8}$, at $n = 5{,}419{,}351$, so every term is computed to about eight correct digits, and the sum is correct to the digits quoted. For much larger $n$ the reduction of $n$ modulo $\pi$ needs more precision than double arithmetic gives, and the terms would have to be computed exactly.

The third number in the comparison is a rational number with an enormous denominator, constructed so that its first several continued-fraction terms grow as described. Over the range computed it behaves like the irrational number it approximates; beyond its denominator it would not. The figure shows the mechanism of divergence, not a proof of divergence for any particular irrational number.

The sum 294.29 for $1/\pi$ is not exactly $\pi^2$ times the Flint Hills sum, because $|\sin n|$ and $\pi\|n/\pi\|$ agree only when both are small; the ordinary terms differ by a bounded factor, and the spikes, which carry almost all of both sums, agree to many digits. The comparison between the three numbers is unaffected, since the divergence of the third comes entirely from its spikes.

And the table of convergents stops at eleven. The later convergents of $\pi$ are known to thousands of terms, and their exponents cluster around 2 with occasional excursions, but no finite table says anything about the supremum.

Still open: whether the series converges #

Whether $\sum 1/(n^3 \sin^2 n)$ converges is unknown, and by Alekseyev’s theorem it is equivalent, up to the boundary case, to whether $\pi$’s irrationality measure is less than $5/2$. The analogous series with a different power, $\sum 1/(n^u \sin^v n)$, runs into the same question at a different threshold, $1 + u/v$ in place of $5/2$, so each choice of powers asks about a different bound on $\pi$’s measure, and all of them are open between their threshold and 7.1.

The deeper open question behind all of them is whether $\pi$ is, in its approximation by fractions, a typical number. Every statistical test says it is. The best proof says only that it is not worse than a number approximated with exponent 7.1, and the distance between those two statements is where the Flint Hills series lives.

A sum decided by one number #

The series was chosen to be easy to state and impossible to decide, and it achieves both with nothing more than a sine and a cube. Its terms are small except at rare moments, the rare moments are the good fractions for $\pi$, and the size of those moments is a property of $\pi$ that is believed on overwhelming evidence and proved by no one. The staircase that looks flat after 355 is flat because $\pi$ has so far behaved like a typical number; whether it stays flat is a question about every fraction that will ever approximate $\pi$, which is more than any computation can see.