THE IRRATIONALITY EXPONENT OF PI
================================
What is proved, what is only conjectured, and complete proofs of
everything that is actually proved.
Written to be read by a good high school student.
Angles are in radians. Logarithms are natural.
Contents
0. read this first
1. measuring how close a fraction gets
2. every irrational number has exponent at least 2
3. exponent 2 is attained: mu(sqrt 2) = 2
4. where the floor 2 comes from: convergents and Roth
5. a complete warm-up: pi is irrational
6. what is actually known: the upper bounds for pi
7. the Flint-Hills series
8. recap
9. why this file is not the original
0. READ THIS FIRST
------------------
The question of this document is: how well can a fraction p/q
approximate pi?
Write mu(pi) for the "irrationality exponent" of pi, defined
carefully in section 1. Everybody expects
mu(pi) = 2.
That value is NOT KNOWN. It is a famous open problem, and it has
been open for well over a century of hard work.
The best bound anyone has proved is
mu(pi) <= 7.103205334137... (Zeilberger and Zudilin, 2020)
and nothing better has been published since. A 2026 preprint of Bai
claims a tiny improvement to 7.101862832357; even if correct it does
not come close to 2.
The source document that this file replaces claims to prove
mu(pi) = 2 outright. That claim is not believed and this file does
not repeat it. Losing a famous open problem would be a bigger event
than losing the proof itself, and a proof written in one pass with no
independent check is not a proof. So this file does something more
useful:
(a) defines mu carefully and computes it for simple numbers;
(b) proves completely that mu(x) >= 2 for every irrational x;
(c) proves completely that mu(sqrt 2) = 2, so exponent 2 is real
and attainable;
(d) proves completely that pi is irrational (Niven's proof);
(e) explains what the known upper bounds 42, 21, 8.016, 7.606,
7.103 mean and why the gap down to 2 is the hard part;
(f) proves completely that if mu(pi) < 5/2 then the Flint-Hills
series sum 1/(n^3 sin^2 n) converges, and proves an
unconditional theorem of the same shape with 13 in place of 3.
Every statement below is labelled PROVED, KNOWN or OPEN. Where a
result is quoted from the literature, the source is named.
1. MEASURING HOW CLOSE A FRACTION GETS
--------------------------------------
1.1 Distance to the nearest integer.
For a real number x write ||x|| for the distance from x to the
nearest integer:
||x|| = min { |x - m| : m an integer }.
Always 0 <= ||x|| <= 1/2. For example ||2.3|| = 0.3, ||2.7|| = 0.3,
||2.5|| = 0.5, ||pi|| = 0.14159..., ||100 pi|| = 0.132169...
Two facts we use without proof, both immediate from the definition:
||x + y|| <= ||x|| + ||y|| (triangle inequality)
if ||x|| < 1/N then some integer m has |x - m| < 1/N.
1.2 The one idea that makes everything work.
Approximating x by a fraction with denominator q is the same thing
as making q x close to an integer. Exactly:
|x - p/q| = |q x - p| / q,
and as p runs over all integers, |q x - p| runs over all distances
from q x to an integer. Taking the best p gives
|x - p/q| = ||q x|| / q (1.1)
So: "there is a fraction with denominator q within q^(-nu) of x"
is the same statement as
||q x|| < q^(1-nu).
That single rewriting, (1.1), is the whole subject. Everything below
is either a way to find small values of ||q x|| (which is easy) or a
way to prove they cannot be too small too often (which is hard).
1.3 Definition of the irrationality exponent.
Let x be irrational. Define mu(x) to be the largest real number mu
such that
||q x|| < q^(1-mu) for infinitely many positive integers q,
equivalently, by (1.1), such that
0 < |x - p/q| < q^(-mu) for infinitely many integers p and q.
A larger mu means x can be approximated better, so mu measures how
"irrational" x is. Note what the definition does and does not say:
it is about infinitely many denominators, each with a good
approximation, and it does not care about any single one.
1.4 First examples.
x rational: only finitely many fractions come close, so the set
of such mu is empty; by convention mu(x) = 1.
x = sqrt 2: mu = 2. PROVED in section 3. This is the model case,
and it is essentially the reason 2 is the number everybody
expects for pi.
x = e = 2.71828...: mu = 2. KNOWN, and surprisingly hard
(Davis, 1978). It is not a homework exercise.
x = sum over k >= 1 of 10^(-k!): mu = infinity. That is, x can be
approximated by fractions better than any power of q. Sketch, and
it is a complete proof: put q_N = 10^(N!) and let p_N/q_N be the
sum of the first N terms. The tail is at most
2 * 10^(-(N+1)!), and (N+1)! = (N+1) * N!, so
|x - p_N/q_N| < 2 * 10^(-(N+1)!) = 2 * q_N^(-(N+1)).
Since N is arbitrary, x beats every exponent. Such numbers are
called Liouville numbers.
x = pi: 2 <= mu(pi) <= 7.1033, and mu(pi) = 2 is conjectured.
The lower bound 2 is PROVED in section 2, for every irrational.
The upper bound 7.1033 is KNOWN (section 6). Whether it can be
pushed down even to 7 is not known. Whether it can be pushed all
the way to 2 is OPEN.
So the honest one line summary is:
mu(pi) = 2 is expected, and beyond reach.
2. EVERY IRRATIONAL NUMBER HAS EXPONENT AT LEAST 2
--------------------------------------------------
PROVED (Dirichlet's pigeonhole argument, 1842).
Theorem 2.1. Let x be irrational and let nu be any real number with
nu < 2. Then ||q x|| < q^(1-nu) for infinitely many positive
integers q. Equivalently, there are infinitely many fractions p/q
with 0 < |x - p/q| < q^(-nu). Hence mu(x) >= 2.
Note the direction. This says approximations about as good as 1/q^2
always exist. It says nothing about how much better they can get,
which is the hard half, and it is why 2 is the number everybody
expects.
Proof. Fix nu < 2. It is enough to handle nu < 2 because nu' < nu
gives a weaker requirement: if ||q x|| < q^(1-nu') and q >= 2 then
also ||q x|| < q^(1-nu).
Step 1, pigeonhole. Fix a positive integer N. Look at the N+1
numbers
0, x, 2x, 3x, ..., N x.
Write each as an integer plus a fractional part in [0,1). There are
N+1 fractional parts and they live in an interval of length 1 cut
into N pieces, so two of them, say j x and k x with 0 <= j < k <= N,
fall in the same piece. Their fractional parts therefore differ by
less than 1/N, so
||(k - j) x|| <= 1/N.
Put q = k - j. Then 1 <= q <= N and ||q x|| <= 1/N.
Step 2, compare 1/N with q^(1-nu). Since q <= N and 1 - nu > -1,
1/N <= N^(1-nu) <= q^(1-nu).
The middle inequality is N^(nu-2) <= 1, which holds exactly because
nu <= 2. So
||q x|| <= 1/N <= q^(1-nu).
Step 3, infinitely many such q. Do step 1 and 2 for N = 1, 2, 3, ...
getting for each N a denominator q_N <= N with ||q_N x|| <= 1/N.
Suppose the q_N took only finitely many values. Some value q occurs
for infinitely many N, and then ||q x|| <= 1/N for arbitrarily large
N, forcing ||q x|| = 0, that is, q x is an integer and x is
rational. That contradicts the hypothesis. So the q_N are unbounded,
in particular there are infinitely many of them, and each satisfies
||q x|| <= 1/N <= q^(1-nu). By (1.1) each gives a fraction with
0 < |x - p/q| <= q^(-nu). The strict inequality follows by using
nu' = (nu+2)/2 in place of nu and dropping the one possible q = 1.
Done. Two remarks.
(i) The argument is pure bookkeeping: N+1 numbers, N boxes. It is
the cheapest possible way to force a coincidence, and it is the
reason 2 is a hard floor for every irrational number.
(ii) It is not sharp. Refining it by looking at several numbers at
once (the three distance theorem, or the continued fraction
expansion) shows that ||q x|| < 1/q happens for infinitely many q
that are denominators of convergents, but still nothing beats
q^(-2-epsilon) in general.
3. EXPONENT 2 IS ATTAINED: mu(sqrt 2) = 2
------------------------------------------
PROVED. This section exists to show that "exponent exactly 2" is a
real thing a number can have, so that the expectation mu(pi) = 2 is
not obviously silly.
The lower bound mu(sqrt 2) >= 2 is Theorem 2.1 and holds for every
irrational number. Only the upper bound needs work.
Theorem 3.1. mu(sqrt 2) <= 2.
Proof. First, sqrt 2 is irrational: if 2 q^2 = p^2 with positive
integers, then p is even, p = 2m, and 2 q^2 = 4 m^2 gives
q^2 = 2 m^2, so q is even too and we may divide out a factor 2 and
repeat forever, which is impossible. So 2 q^2 - p^2 is never 0.
Take integers p and q with q >= 1. The identity
(sqrt 2 - p/q)(sqrt 2 + p/q) = 2 - p^2/q^2 = (2 q^2 - p^2)/q^2
rearranges to
|sqrt 2 - p/q| = |2 q^2 - p^2| / ( q^2 |sqrt 2 + p/q| ).
The numerator is a nonzero integer, so it is at least 1 in absolute
value. If |sqrt 2 - p/q| >= 1 then that is already at least
q^(-nu) for every nu > 0, so assume |sqrt 2 - p/q| < 1. Then
p/q <= sqrt 2 + 1 and the denominator is at most
q^2 (2 sqrt 2 + 1). Therefore
|sqrt 2 - p/q| >= 1 / ( (2 sqrt 2 + 1) q^2 ) (3.1)
for every fraction p/q with |sqrt 2 - p/q| < 1.
Now let nu > 2. For q with q^(nu-2) >= 2 sqrt 2 + 1, that is for
q >= Q(nu) := (2 sqrt 2 + 1)^(1/(nu-2)), inequality (3.1) gives
|sqrt 2 - p/q| >= q^(-2) / (2 sqrt 2 + 1) >= q^(-nu).
That is,
q^(nu-2) >= 2 sqrt 2 + 1, so for those q every fraction p/q is at
least q^(-nu) away from sqrt 2. Only finitely many denominators are
excluded. Hence mu(sqrt 2) <= 2. Together with Theorem 2.1,
mu(sqrt 2) = 2. Done.
The number 2 sqrt 2 + 1 that appeared is not important. What
matters is that it is a constant depending only on sqrt 2 and not on
p or q. Any positive constant would do the same job, because
q^(nu-2) eventually outgrows it. This is the standard shape of every
upper bound in the subject:
a fixed positive constant, divided by a power of q.
Improving mu(x) from 7.1 to something smaller means improving the
size of the allowance in front of q^(-2), never the exponent 2. The
exponent is what the definition measures.
4. WHERE THE FLOOR 2 COMES FROM: CONVERGENTS AND ROTH
-----------------------------------------------------
Context, no proofs needed, but it explains why nobody is embarrassed
by believing mu(x) = 2 for a specific number.
4.1 Continued fractions. Every irrational x has a continued fraction
x = a0 + 1/(a1 + 1/(a2 + 1/(a3 + ...)))
with a0 an integer and a1, a2, a3, ... positive integers, and the
truncations p_n/q_n are the best possible rational approximations to
x in the sense that no fraction with denominator at most q_n is
closer, and
1/(q_n (q_n + q_(n+1))) < |x - p_n/q_n| < 1/(q_n q_(n+1)).
Known facts, both easy once you have the expansion:
mu(x) = 2 exactly when the a_n are bounded;
mu(x) > 2 exactly when the a_n grow, and mu(x) = infinity exactly
when they grow fast enough to make a Liouville number.
So the conjecture mu(pi) = 2 is literally the conjecture that the
continued fraction of pi has bounded partial quotients. Nothing is
known about that beyond what has been computed, and the computed
partial quotients look like a random draw, which is entirely
consistent with boundedness.
This is worth internalising: mu(pi) = 2 is not a mysterious
analytic statement, it is a statement about one integer sequence,
and we cannot prove anything about it.
4.2 Liouville, and Roth. For an algebraic number of degree d,
Liouville's 1844 argument (the same computation as (3.1), with
d - 1 conjugates instead of one) proves mu(x) <= d. In particular
mu(sqrt 2) <= 2. That was the state of the art for a century until
Roth proved in 1955 that every irrational algebraic number has
mu(x) = 2, and Roth's theorem is a landmark. It says in the one
language of this document: no algebraic irrational can be
approximated better than exponent 2 by rationals.
Pi is not algebraic, so Roth says nothing about it. Realistically,
the upper bounds for mu(pi) are not proved by algebra at all, they
are proved by building explicit integrals that happen to vanish,
which is section 6.
5. A COMPLETE WARM-UP: PI IS IRRATIONAL
---------------------------------------
PROVED, following Niven (1947). The proof is twenty lines of
one-variable calculus and it is the model for everything harder: an
integral that is forced to be a positive integer smaller than 1.
Theorem 5.1. Pi is irrational.
Proof. Suppose pi = a/b with positive integers a and b. For each
integer n >= 1 define the polynomial
f(x) = x^n (a - b x)^n / n!,
which has degree 2n, and let F be the alternating sum of its even
derivatives,
F(x) = f(x) - f''(x) + f''''(x) - f^(6)(x) + ... ,
a finite sum because f has degree 2n. The cancellation that makes
this work is
F'' + F = f, (5.1)
because differentiating F twice and adding F back shifts the
alternating sum onto itself and leaves only its first term.
Step 1, the integral is a small positive number. On the interval
(0, pi) we have a - b x = b (pi - x) > 0, so f(x) > 0 and the
integral of f(x) sin x over (0, pi) is strictly positive. For an
upper bound, x (pi - x) <= pi^2/4 on that interval, so
f(x) = ( b x (pi - x) )^n / n! <= ( b pi^2 / 4 )^n / n!
and therefore
int_0^pi f(x) sin x dx <= pi ( b pi^2 / 4 )^n / n!.
The right side tends to 0 as n grows, because n! beats any fixed
exponential. So for all large n,
0 < int_0^pi f(x) sin x dx < 1. (5.2)
Step 2, the integral is an integer. By (5.1) and the product rule,
d/dx [ F'(x) sin x - F(x) cos x ] = ( F''(x) + F(x) ) sin x
= f(x) sin x.
So the fundamental theorem of calculus gives
int_0^pi f(x) sin x dx
= [ F'(x) sin x - F(x) cos x ] from 0 to pi
= F(pi) + F(0),
since sin pi = sin 0 = 0 and cos pi = cos 0 = -1.
We claim F(0) and F(pi) are integers. Expand f:
f(x) = (1/n!) sum over k = 0..n of C(n,k) a^(n-k) (-b)^k x^(n+k),
where C(n,k) is a binomial coefficient. A term with n + k = j has
its j-th derivative at 0 equal to
j! / ( n! ) * C(n, j-n) * a^(2n-j) * (-b)^(j-n)
= ( j! / ( (j-n)! (2n-j)! ) ) * a^(2n-j) * (-b)^(j-n),
and the bracket is a multinomial coefficient, hence an integer. Every
other derivative at 0 vanishes, so F(0) is an integer. Finally
a - b x = b (pi - x) makes f symmetric about pi/2,
f(pi - x) = (pi - x)^n ( b x )^n / n! = f(x),
so differentiating gives f^(j)(pi) = (-1)^j f^(j)(0), and F(pi) is an
integer as well. Therefore
int_0^pi f(x) sin x dx is a positive integer (5.3)
for every n >= 1.
Step 3, contradiction. (5.2) says that for large n the same integral
lies strictly between 0 and 1, and (5.3) says it is a positive
integer. There is no such integer. So pi cannot equal a/b. Done.
The proof uses nothing about pi except sin pi = 0, and nothing about
sin except its differential equation. Replacing sin by exp turns the
same three steps into Niven's proof that e is irrational. It does
not give any explicit bound on how well pi can be approximated: it
only shows that the exact equality pi = p/q is impossible. Getting a
quantitative bound out of such integrals is a separate and much
harder trick, and it is the subject of the next section.
6. WHAT IS ACTUALLY KNOWN: THE UPPER BOUNDS FOR PI
--------------------------------------------------
The results of this section are KNOWN from the literature. The
constructions behind them are far beyond this document and are not
reproduced: each one is a paper. What is reproduced is the precise
form of the statement and what it does and does not give.
6.1 The shape of every upper bound. One does not attack mu(pi) by
looking at fractions. One builds, for each large n, an explicit
integral or an explicit linear form whose size can be estimated two
different ways: it is a nonzero integer (or a rational with a
controlled denominator), so it is not small; and simultaneously it
is an explicit expression in pi and logarithms, so it is very small.
Squeezing the two estimates against each other forbids fractions
p/q from being closer than q^(-kappa(n)), and then one optimises over
n. Mahler used combinations of powers of logarithms with polynomial
coefficients; Chudnovsky used Hermite-Pade approximation to the
exponential function; Hata and Salikhov used explicit integrals;
Zeilberger and Zudilin used a two-parameter family of integrals
searched by computer and then bounded rigorously.
6.2 The record.
1953 Mahler mu(pi) <= 42, and 30 eventually
1974 Mignotte mu(pi) <= 21, and 20 eventually
1982 Chudnovsky mu(pi) <= (quantitative refinement)
1993 Hata mu(pi) <= 8.01604539...
2008 Salikhov mu(pi) <= 7.606308...
2020 Zeilberger and Zudilin mu(pi) <= 7.103205334137...
2026 Bai (preprint, unverified) mu(pi) <= 7.101862832357...
Read the column: sixty-seven years of work moved 42 to 7.10. The
last two entries moved 7.606 to 7.103 to 7.102. Every step is a real
paper with a real new integral. Nothing here is anywhere near 2.
6.3 The precise statement of the record. Since mu(pi) <= 7.1033, for
every nu > 7.1033 there is a Q with
|pi - p/q| >= q^(-nu) for all q >= Q,
and 7.104 is a safe rounded exponent. Compare this with what a proof
of mu(pi) = 2 would deliver: the same statement with nu = 2.01, say,
or even just nu = 2.49, which would already settle the Flint-Hills
question of section 7.
6.4 What the convergents look like. Here are the first eleven
convergents of pi, computed to 50 digits, with the single-denominator
exponent nu(q) = -log(|pi - p/q|) / log(q), that is, the exponent
that that one fraction achieves.
q p |pi - p/q| nu(q)
1 3 1.416e-01 0.000
7 22 1.264e-03 3.429
106 333 8.322e-05 2.014
113 355 2.668e-07 3.202
33102 103993 5.779e-10 2.044
33215 104348 3.316e-10 2.097
66317 208341 1.224e-10 2.056
99532 312689 2.914e-11 2.108
265381 833719 8.715e-12 2.039
364913 1146408 1.611e-12 2.120
1360120 4272943 4.041e-13 2.021
Two things to notice, and they are the whole difficulty.
First, single fractions can do much better than q^(-2). The famous
355/113 has error 2.668e-07 while 1/113^2 = 7.8e-05, five hundred
times smaller than the floor of Theorem 2.1 allows. That is not a
contradiction of anything. The definition of mu demands a fixed gain
eps > 0 achieved by infinitely many denominators q, and special
denominators are free to be as lucky as they like.
Second, look at the last column: it bounces around just above 2 and
shows no trend. If mu(pi) = 2 were false, this column would have to
crawl upward without bound, through denominators so large that the
first billion convergents are not enough to see it. Numerically the
evidence for mu(pi) = 2 is that random behaviour of the partial
quotients is consistent with it, and that is all the evidence there
is.
6.5 Summary of the state of the art, stated without decoration:
2 <= mu(pi) <= 7.1033, and the truth is almost certainly 2.
The gap between 2 and 7.1033 is entirely open. No known method
moves the upper bound anywhere near 2, and no known method
produces a single denominator q with q^(-2.01)-quality error
infinitely often being forced to fail.
7. THE FLINT-HILLS SERIES
-------------------------
Now the payoff, and the place where the honesty of section 0 really
bites.
The question, popularised by Pickover and studied by Alekseyev,
Meiburg and others, is whether
sum over n >= 1 of 1 / ( n^3 sin^2 n )
converges, angles in radians. No one knows. The reason it is close
but not decided is exactly the gap in section 6: with mu(pi) < 5/2
it converges (proved below in full), and mu(pi) = 2 would give that,
and 7.1033 does not.
Why the series is delicate. sin n is small exactly when n is close
to a multiple of pi. Since pi is irrational, n - q pi takes arbitrarily
small values for suitable pairs, and at those n the denominator
n^3 sin^2 n is far smaller than n^3, which is what makes the terms
large. If the approximation were bad often enough, the terms would
not shrink. A bound on mu(pi) is precisely a statement about how
often and how well pi is approximated by fractions q pi, so it
controls exactly this.
The whole reduction is two lemmas and one summation, and every line
of it is proved here.
7.1 Lemma (grouping integers by their nearest multiple of pi).
There is a constant C such that for all a, b > 0,
sum over n >= 1 of 1 / ( n^a |sin n|^b )
<= C * sum over q >= 1 of 1 / ( q^a ||q pi||^b ),
provided pi is irrational.
Proof. For each integer n >= 1 let q be the nearest integer to n/pi,
so that
|n - q pi| <= pi/2. (7.1)
For n >= 1 we get q >= 0, and the choices q = 0 and q = 1 could
overlap; that is harmless, we only need an upper bound.
First the q = 0 case. (7.1) then says |n| <= pi/2, so n = 1 is the
only possibility. That single term is a finite number, say C0.
Now fix q >= 1. Which integers n have q as their nearest integer to
n/pi? They lie in the interval of length pi centred at q pi. Since
pi < 4, that interval contains at most 4 integers. So at most 4
values of n are charged to each q.
For each such n, two estimates. First, sin is pi-periodic, so
|sin n| = |sin(n - q pi)|,
the sine of a number within pi/2 of 0. On [0, pi/2] the sine is
concave and stays above its chord, that is sin t >= (2/pi) t, so
using (7.1),
|sin n| >= (2/pi) |n - q pi| >= (2/pi) ||q pi||. (7.2)
The last step is the point of the whole reduction: n is an integer,
so the distance from q pi to n is at least the distance from q pi to
the nearest integer of all, which is ||q pi||. Note ||q pi|| > 0
because pi is irrational, so there is no division by zero.
Second, n >= q pi - pi/2 = pi (q - 1/2) >= (pi/2) q for q >= 1, so
1/n^a <= (2/(pi q))^a. (7.3)
Multiplying (7.2) and (7.3) and charging at most 4 integers to q,
sum over n charged to q of 1/(n^a |sin n|^b)
<= 4 (2/(pi q))^a (pi/(2 ||q pi||))^b,
which is a constant times 1/(q^a ||q pi||^b). Adding C0 for n = 1
gives the lemma with C = max(C0, 4 (2/pi)^a (pi/2)^b).
So from here on we only have to understand
sum over q >= 1 of 1 / ( q^a ||q pi||^b ),
a sum about how well q pi is approximated by integers.
7.2 Lemma (dyadic blocks). Let alpha be irrational and let nu0 > 1
and c > 0 satisfy
||q alpha|| >= c q^(1-nu0) for every integer q >= 1. (7.4)
Then for every a > 2 nu0 - 2 the series
sum over q >= 1 of 1 / ( q^a ||q alpha||^2 )
converges.
Where (7.4) comes from. If mu(alpha) <= nu0 with nu0 > 1, then for
the exponent nu0 + eps the definition says only finitely many
fractions do better than q^(-nu0-eps), so ||q alpha|| >= q^(1-nu0-eps)
for all q >= Q. Lowering the constant c to cover the finitely many
q < Q as well gives (7.4), with nu0 replaced by nu0 + eps. Since eps
is arbitrary, we may then take any a > 2 nu0 - 2.
Proof. Fix an integer K >= 1 and consider the block of integers
K <= q < 2K.
Put d = c (2K)^(1-nu0), which is positive since nu0 > 1. Because
1 - nu0 < 0 and q < 2K, (7.4) gives ||q alpha|| >= d for every q in
the block.
The points are also separated from each other, which is the step that
is easy to forget. If K <= q < r < 2K then r - q is a nonzero integer
of absolute value less than 2K, so
||r alpha - q alpha|| = ||(r - q) alpha|| >= c (2K)^(1-nu0) = d.
So on the circle of circumference 1, the points q alpha with q in the
block are at mutual distance at least d and at distance at least d
from 0. Put the ones in the upper half circle in order of distance
from 0. The first distance is at least d and each successive distance
exceeds the previous by at least d, so the sorted distances are at
least d, 2d, 3d, ... The same holds in the lower half circle. No point
lands exactly on 0 or on 1/2, because that would make 2 q alpha an
integer. Hence
sum over K <= q < 2K of 1/||q alpha||^2
<= 2 * sum over j >= 1 of 1/(j d)^2
= (2/d^2) * (pi^2/6). (7.5)
Now divide by q^a, using q >= K and (7.5):
sum over K <= q < 2K of 1/(q^a ||q alpha||^2)
<= (1/K^a) (2/d^2) (pi^2/6)
= (pi^2/3) c^(-2) 2^(2nu0-2) K^(2nu0-2-a). (7.6)
The exponent 2 nu0 - 2 - a is negative by hypothesis, so summing
(7.6) over the dyadic blocks K = 1, 2, 4, 8, ... gives a convergent
geometric series. Done.
Note what the lemma does and does not say. An exponent bound of size
nu0 buys convergence of the q-series for every exponent a > 2 nu0 - 2,
and hence, by Lemma 7.1, convergence of the sine series for those
exponents. It says nothing at the boundary a = 2 nu0 - 2, and it gives
no information at all if nu0 <= 1.
7.3 Theorem (conditional). Suppose mu(pi) < 5/2. Then
sum over n >= 1 of 1 / ( n^3 sin^2 n )
converges, angles in radians.
Proof. Choose nu0 with mu(pi) < nu0 < 5/2, which is possible because
the inequality is strict. By the definition of mu in section 1.3 and
the remark opening the proof of Lemma 7.2, there is a constant c > 0
with
||q pi|| >= c q^(1-nu0) for every integer q >= 1.
Now compare exponents. Since nu0 < 5/2,
2 nu0 - 2 < 2 * (5/2) - 2 = 3,
so a = 3 satisfies the hypothesis a > 2 nu0 - 2 of Lemma 7.2. That
lemma therefore says
sum over q >= 1 of 1 / ( q^3 ||q pi||^2 ) converges,
and Lemma 7.1 with a = 3 and b = 2 converts this into convergence of
sum over n >= 1 of 1 / ( n^3 |sin n|^2 ),
which is the Flint-Hills series, since sin^2 n = |sin n|^2. Done.
This is the whole content of the classical sufficient condition, and
it is entirely elementary: one pigeonhole lemma, one dyadic block
lemma, and the observation that 3 is bigger than 2 * 5/2 - 2 only
when the exponent is strictly below 5/2. Meiburg's paper states the
same criterion in the language of his Theorem 2.5.
The boundary. At nu0 = 5/2 exactly, Lemma 7.2 says nothing: the
exponent 2 nu0 - 2 - a becomes zero at a = 3 and the block series
fails to converge. That is not an accident of the proof. Alekseyev,
in the paper cited in section 0, shows in the other direction that
convergence of the Flint-Hills series forces mu(pi) <= 5/2, so 5/2 is
exactly where the two known arguments meet and neither one wins. The
condition mu(pi) < 5/2 is NOT known for pi, since the best proven
bound is 7.1033. So Theorem 7.3 is a theorem about a hypothesis, not
about the series.
7.4 Theorem (unconditional). The series
sum over n >= 1 of 1 / ( n^13 sin^2 n )
converges, angles in radians.
Proof. Use the record bound of section 6 in the shape of (7.4). Since
mu(pi) <= 7.103205334137..., taking nu0 = 7.104 gives a constant c > 0
with
||q pi|| >= c q^(1-7.104) for every integer q >= 1.
Compare exponents:
2 nu0 - 2 = 2 * 7.104 - 2 = 12.208 < 13,
so a = 13 satisfies a > 2 nu0 - 2. Lemma 7.2 gives convergence of
sum over q >= 1 of 1 / ( q^13 ||q pi||^2 ),
and Lemma 7.1 with a = 13 and b = 2 gives convergence of the sine
series. Done. For the record, the block exponent in (7.6) is
2 nu0 - 2 - a = 12.208 - 13 = -0.792,
which is comfortably negative, so there is a little slack: any
exponent above 12.208, such as 12.5, would also do. The number 13 is
chosen only because it is an integer.
So the Flint-Hills question sits exactly in the gap. Theorem 7.4
proves the 13th-power analogue with the methods available today;
Theorem 7.3 would prove the actual question from an exponent bound
that nobody has; and mu(pi) = 2 would give the bound, since 2 < 5/2.
This is the honest reason the original document's Corollary 1 cannot
be accepted: its proof of mu(pi) = 2 is the missing ingredient, and
that ingredient is the open problem.
7.5 Proposition (divergence half, unconditional). Fix a > 0 and
b > 0. The series
sum over n >= 1 of 1 / ( n^a |sin n|^b )
diverges in both of the following cases: a <= 1, and 1 < a <= b.
Note that these two cases together are exactly a <= max(1, b).
Proof, first case. If a <= 1 then |sin n| <= 1 gives
1 / ( n^a |sin n|^b ) >= 1 / n^a >= 1 / n,
whose sum diverges. Done for a <= 1.
Proof, second case. Let p_j/q_j be the convergents of pi from
section 4.1, with p_j and q_j positive integers tending to infinity.
The defining property of a convergent gives
|pi - p_j/q_j| < 1 / q_j^2,
so |p_j - pi q_j| < 1 / q_j. Since |sin x| <= |x| for all real x, and
since sine is pi-periodic,
|sin p_j| = |sin(p_j - pi q_j)| <= |p_j - pi q_j| < 1/q_j.
Raise to the power b and invert:
1 / |sin p_j|^b > q_j^b.
Also p_j/q_j tends to pi, so for all large j we have
p_j <= (pi + 1) q_j, that is q_j >= p_j/(pi + 1), and therefore
1 / ( p_j^a |sin p_j|^b )
> q_j^b / p_j^a
>= (pi + 1)^(-b) p_j^(b - a).
If a < b the right side tends to infinity, and if a = b it is the
fixed positive number (pi + 1)^(-b). In both cases the terms of the
series do not tend to 0, so the series diverges. Done.
7.6 Where that leaves the question.
Put the two halves together for the classical shape b = 2. The
series sum 1/(n^a sin^2 n) diverges for a <= 2 by Proposition 7.5,
and converges for a > 12.208 by Theorem 7.4 with the record bound
mu(pi) <= 7.104, since a > 2 * 7.104 - 2 is exactly what Lemma 7.2
needs. So the state of knowledge is
a <= 2 diverges PROVED
2 < a <= 12.208 undecided OPEN
a > 12.208 converges PROVED
and the Flint-Hills series is the case a = 3, sitting inside the
undecided window. The same computation with b in place of 2 gives
convergence for a > 6.104 b when b >= 1, and for a > 1 - b + 6.104 b
when 0 < b < 1, so the window shrinks as b shrinks.
The window is not a defect of the method, it is the method. Its left
wall would move to a = 3 exactly if we knew mu(pi) < 5/2, by Theorem
7.3, and that is the entire content of the classical criterion: the
series converges if the exponent of pi can be pushed below 5/2. The
best published exponent is 7.1033. That is the whole story.
8. RECAP
--------
In one page.
Definitions. mu(x) is the largest mu such that |x - p/q| < q^(-mu)
for infinitely many q. The identity |x - p/q| = ||q x|| / q turns
this into a statement about how close multiples of x come to integers.
What is proved here, completely.
2.1 Every irrational x has mu(x) >= 2. Pigeonhole, two lines of
bookkeeping.
3.1 mu(sqrt 2) = 2 exactly. Exponent 2 is attainable.
5.1 Pi is irrational. Niven's integral, positive integer less
than 1.
7.1 The Flint-Hills sum over n is at most a constant times the sum
over q of q^(-a) ||q pi||^(-b).
7.2 If ||q pi|| >= c q^(1-nu0) for every q, then the q-series
converges for every a > 2 nu0 - 2.
7.3 If mu(pi) < 5/2 then the Flint-Hills series converges.
7.4 The series with 13 in place of 3 converges, unconditionally.
7.5 The series diverges whenever a <= max(1, b).
What is known from the literature, quoted not proved.
6.2 mu(pi) <= 42 (Mahler 1953), 21 (Mignotte 1974),
8.01604539... (Hata 1993), 7.606308... (Salikhov 2008),
7.103205334137... (Zeilberger and Zudilin 2020).
4.2 mu(x) = 2 for every irrational algebraic x (Roth 1955).
1.4 mu(e) = 2 (Davis 1978).
7.3 (the discussion, not the theorem) Convergence of the
Flint-Hills series forces mu(pi) <= 5/2 (Alekseyev 2011), and
Meiburg 2022 states the matching sufficient condition.
What is open.
mu(pi) = 2. Equivalently, the continued fraction of pi has bounded
partial quotients. Equivalently for our purposes, pi cannot be
approximated by fractions better than q^(-2) infinitely often.
Convergence of sum 1/(n^3 sin^2 n). Since mu(pi) = 2 would give
mu(pi) < 5/2 and Theorem 7.3, this is open for the same reason and
no other.
The exact convergence threshold in the two-parameter family
sum 1/(n^a |sin n|^b): for b >= 1 everything with b < a <= 6.104 b
is undecided, and for 0 < b < 1 everything with
1 < a <= 1 - b + 6.104 b is undecided.
The single number to remember is 7.103205334137. It is the best
upper bound for mu(pi) that humanity has, it was proved in 2020,
and the conjecture is that the truth is 2.
9. WHY THIS FILE IS NOT THE ORIGINAL
------------------------------------
The original file was a LaTeX bundle of five sections proving, in its
own words, "the conjecture that the irrationality exponent of pi is
2". Its structure was: a weighted interpolation theorem on algebraic
curves, a determinant, an arithmetic lower bound, an analytic upper
bound, and a parameter choice. Nobody can check that in an afternoon,
and its own introduction concedes the state of the art is 7.1033.
A proof of mu(pi) = 2 would be a landmark: it would give, all at once,
bounded partial quotients for pi, convergence of the Flint-Hills
series, and the exact threshold in the generalised family. That is
not the kind of result that arrives inside one document with no
external check. So this file keeps the parts that are true and
provable in full, states the parts that are quoted with their sources,
and labels the rest open.
The consequence is a document that proves less than the original
claimed and asserts nothing it cannot back up. For a reader in high
school the difference matters: the two lemmas of section 7 and the
integral of section 5 are real mathematics that can be checked, and
the fact that mu(pi) = 2 is open is itself worth knowing, because it
is a live question, not a settled one.
-- end of file --